Algebraic Identities
1. Difference of Two Squares:
\[
a^2 – b^2 = (a – b)(a + b)
\]
Examples:
\( \bullet \quad x^2 – y^4 = x^2 – (y^2)^2 = (x – y^2)(x + y^2) \)
\( \bullet \quad x – y^2 = (\sqrt{x})^2 – y^2 = (\sqrt{x} – y)(\sqrt{x} + y) \)
\( \bullet \quad x^{2a} – 1 = (x^a)^2 – 1^2 = (x^a – 1)(x^a + 1) \)
\( \bullet \quad 4^x – 4^{-x} = (2^2)^x – (2^2)^{-x} = (2^x)^2 – (2^{-x})^2 \)
\[
= (2^x – 2^{-x}) (2^x + 2^{-x})
\]
2. Difference of Two Cubes:
\[ a^3 – b^3 = (a – b) (a^2 + ab + b^2) \]
Examples:
\( \bullet \quad 8^x – (27)^x = (2^3)^x – (3^3)^x = (2^x)^3 – (3^x)^3 \)
\[
= (2^x – 3^x)(2^{2x} + 2^x \cdot 3^x + 3^{2x})
\]
\[
= (2^x – 3^x)(4^x + 6^x + 9^x)
\]
\( \bullet \quad x^6 – y^6 = (x^2)^3 – (y^2)^3 \)
\[
= (x^2 – y^2)(x^4 + x^2y^2 + y^4)
\]
\[
= (x – y)(x + y)(x^4 + x^2y^2 + y^4)
\]
3. Sum of Two Cubes:
\[
a^3 + b^3 = (a + b) (a^2 – ab + b^2)
\]
Examples:
\( \bullet \quad x^3 + 1 = x^3 + 1^3 = (x + 1)(x^2 – x + 1) \)
\( \bullet \quad 8x^3 + 125 = (2x)^3 + 5^3 \)
\[
= (2x + 5)(4x^2 – 10x + 25) \]
4. General Difference of $n$-th Powers (where $n \in \mathbb{Z^+}$):
\[ a^n – b^n = (a – b)(a^{n-1} + a^{n-2}b + a^{n-3}b^2 + \dots + b^{n-1}) \]
Examples:
\( \bullet \quad 16 – x^4 = 2^4 – x^4 = (2 – x)(2^3 + 2^2x + 2x^2 + x^3) \)
\[
= (2 – x)(8 + 4x + 2x^2 + x^3)
\]
\( \bullet \quad x^7 – 1 = (x – 1)(x^6 + x^5 + x^4 + x^3 + x^2 + x + 1) \)
5. General Sum of $n$-th Powers (where $n \in \mathbb{Z}^+$ and $n$ is an odd integer):
By substituting \( -b \) in place of \( b \) in the fourth identity, we obtain:
\[
a^n + b^n = (a + b)(a^{n-1} – a^{n-2}b + a^{n-3}b^2 – \dots + b^{n-1})
\]
Example:
\[
32 + x^5 = 2^5 + x^5 = (2 + x) (2^4 – 2^3 x + 2^2 x^2 – 2x^3 + x^4)
\]
\[
= (2 + x) (16 – 8x + 4x^2 – 2x^3 + x^4)
\]
6. Square of a Binomial Sum:
\[
(a + b)^2 = a^2 + 2ab + b^2
\]
Examples:
\( \bullet \quad \left( x + \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} + 2 \cdot x \cdot \frac{1}{x} \)
\[
= x^2 + \frac{1}{x^2} + 2
\]
\( \bullet \quad(\sqrt{x} + x)^2 = (\sqrt{x})^2 + x^2 + 2 \cdot \sqrt{x} \cdot x \)
\[
= x + x^2 + 2x\sqrt{x}
\]
7. Square of a Binomial Difference:
\[
(a – b)^2 = a^2 – 2ab + b^2
\]
Examples:
\( \bullet \quad (2^x – 3^x)^2 = (2^x)^2 + (3^x)^2 – 2 \cdot 2^x \cdot 3^x \)
\[
= 4^x + 9^x – 2 \cdot 6^x
\]
\( \bullet \quad \left( \sqrt{x} – \frac{1}{\sqrt{x}} \right)^2 = (\sqrt{x})^2 + \left(\frac{1}{\sqrt{x}}\right)^2 – 2 \cdot \sqrt{x} \cdot \frac{1}{\sqrt{x}} \)
\[
= x + \frac{1}{x} – 2
\]
Note on Perfect Square Trinomials:
You can verify whether a given trinomial expression is a perfect square by using the following visual relationship:
\[
\begin{array}{c l l l c}
& a^2 &\pm & 2ab & + & b^2 & = &(a \pm b)^2 \\
& \downarrow &\downarrow& & \swarrow & & \\
( & a &\pm & \phantom{a} \phantom{a} & b )^2 & \\
&\searrow && \swarrow & \\
&&2ab
\end{array}
\]
Examples:
\( \bullet \quad 9x^2 – 30x + 25 = (3x – 5)^2 \)
\[
\begin{array}{c c }
9x^2& – \;\; 30x & + 25\\
\downarrow &\quad \quad& \downarrow\\
(3x & -& 5)^2\\
\searrow &\quad \quad & \swarrow &\\
&2 \cdot 3x \cdot 5 = 30x\\
\end{array}\]
\( \bullet \quad x^2 + x + \frac{1}{4} = \left( x + \frac{1}{2} \right)^2 \)
\[
\begin{array}{c c }
x^2 & + x& +\frac{1}{4} \\
\downarrow &\quad \quad& \downarrow\\
(x & + &\frac{1}{2} )^2\\
\searrow &\quad \quad & \swarrow &\\
&2 \cdot x \cdot \frac{1}{2} = x\\
\end{array}
\]
\( \bullet \quad 4^x – 2^{x+2} + 4 = (2^x)^2 – 4 \cdot 2^x + 4 = (2^x – 2)^2 \)
\[
\begin{array}{c c }
4^x & – 2^{x+2} &+ 4 \\
\downarrow &\quad \quad& \downarrow\\
(2^x & -& 2)^2\\
\searrow &\quad \quad & \swarrow &\\
&2 \cdot 2^x \cdot 2 = 4 \cdot 2^x\\
\end{array}
\]
\( \bullet \quad a + 6 \sqrt{ a} + 9 = ( \sqrt{ a} + 3 )^2 \)
\[
\begin{array}{c c }
a &+ 6 \sqrt{ a}& + 9 \\
\downarrow &\quad \quad& \downarrow\\
(\sqrt{a } & +& 3)^2\\
\searrow &\quad \quad & \swarrow &\\
&2 \cdot \sqrt{a } \cdot 3 = 6 \cdot \sqrt{a } \\
\end{array}
\]
8. Square of a Trinomial:
\[
(a + b + c)^2 = a^2 + b^2 + c^2 + 2 (ab + ac + bc)
\]
Note:
The bracketed expression on the right-hand side represents twice the sum of all possible pairwise products of the terms in the trinomial, keeping their respective signs.
Examples:
\( \bullet (a – b + c)^2 = a^2 + b^2 + c^2 + 2 (ac – ab – bc) \)
\( \bullet \left( x + 2 \;-\; \frac{1}{x} \right)^2 = x^2 + 4 + \frac{1}{x^2} + 2 \left(2x\; – \;x \cdot \frac{1}{x} \;- \;2 \cdot \frac{1}{x}\right) \)
\[
= x^2 + 4 + \frac{1}{x^2} + 2 \left(2x\; – \;1 \;- \;\frac{2}{x}\right)
\]
9. Cube of a Binomial Sum:
\[
(a + b)^3 = a^3 + 3a^2 b + 3ab^2 + b^3
\]
Examples:
\( \bullet \quad (2 + a)^3 = 2^3 + 3 \cdot 2^2 \cdot a + 3 \cdot 2 \cdot a^2 + a^3 \)
\[
= 8 + 12a + 6a^2 + a^3
\]
\( \bullet \quad 8^a + 3 \cdot 4^a + 3 \cdot 2^a + 1 = (2^a)^3 + 3(2^a)^2 + 3(2^a) + 1 \)
\[
= (2^a + 1)^3
\]
\( \bullet \quad x\sqrt{x} + 3x\sqrt{y} + 3\sqrt{x}y + y\sqrt{y} \)
\[
= (\sqrt{x})^3 + 3(\sqrt{x})^2 \sqrt{y} + 3\sqrt{x} (\sqrt{y})^2 + (\sqrt{y})^3
\]
\[
= (\sqrt{x} + \sqrt{y})^3
\]
10. Cube of a Binomial Difference:
\[
(a – b)^3 = a^3 – 3a^2 b + 3ab^2 – b^3 \]
Examples:
\( \bullet (x – 1)^3 = x^3 – 3x^2 + 3x – 1 \)
\( \bullet 8^a – 3 \cdot (12)^a + 3 \cdot (18)^a – (27)^a \)
\[
= (2^3)^a – 3 \cdot 4^a \cdot 3^a + 3 \cdot 2^a \cdot 9^a – (3^3)^a
\]
\[
= (2^a)^3 – 3(2^a)^2 \cdot 3^a + 3(2^a) (3^a)^2 – (3^a)^3
\]
\[
= (2^a – 3^a)^3
\]
Note on Binomial Expansion and Pascal’s Triangle:
When expanding an expression of the form \((a + b)^n\), the numerical coefficients are obtained using Pascal’s Triangle as shown below:
\[
\begin{array}{c}
n = 0 \quad \quad \quad \quad \quad 1 \\
n = 1 \quad \quad \quad \quad \quad 1 \quad 1 \\
n = 2 \quad \quad \quad \quad \quad 1 \quad 2 \quad 1 \\
n = 3 \quad \quad \quad \quad \quad 1 \quad 3 \quad 3 \quad 1 \\
n = 4 \quad \quad \quad \quad \quad 1 \quad 4 \quad 6 \quad 4 \quad 1 \\
n = 5 \quad \quad \quad \quad \quad 1 \quad 5 \quad 10 \quad 10 \quad 5 \quad 1 \\
n = 6 \quad \quad \quad \quad \quad 1 \quad 6 \quad 15 \quad 20 \quad 15 \quad 6 \quad 1\\
\cdots
\end{array}
\]
In the expansion of \( (a + b)^n \), the algebraic terms corresponding to these coefficients are ordered sequentially as:
\[
a^n, a^{n-1}b, a^{n-2}b^2, a^{n-3}b^3, \dots, ab^{n-1}, b^n
\]
For the expansion of \((a – b)^n\), the signs alternate, such that any term containing an odd power of \( b \) is assigned a negative sign (-).
11. Fourth Power of a Binomial Sum:
\[
(a + b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4
\]
Example:
\[
(2 + x)^4 = 2^4 + 4 \cdot 2^3 \cdot x + 6 \cdot 2^2 \cdot x^2 + 4 \cdot 2 \cdot x^3 + x^4
\]
\[
= 16 + 32x + 24x^2 + 8x^3 + x^4
\]
12. Fourth Power of a Binomial Difference (Substituting \( -b \) into Identity 11):
\[
(a – b)^4 = a^4 – 4a^3b + 6a^2b^2 – 4ab^3 + b^4
\]
13. Fifth Power of a Binomial Sum:
\[
(a + b)^5 = a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5
\]
14. Fifth Power of a Binomial Difference (Substituting \( -b \) into Identity 13):
\[
(a – b)^5 = a^5 – 5a^4b + 10a^3b^2 – 10a^2b^3 + 5ab^4 – b^5
\]
QUESTION 6
Which of the choices below represents a factor of the algebraic expression:
\[
1 – a^2 – 2ax – x^2
\]
\[
\text{A)} 1-a-x \quad
\text{B) } 1-a \quad
\text{C) } x-a \quad
\text{D) } 1-a+x \quad
\text{E) } x+a
\]
Solution:
Group the last three terms and factor out the negative sign to reveal a perfect square trinomial:
\[
1 – a^2 – 2ax – x^2 = 1 – (a^2 + 2ax + x^2)
\]
\[
= 1^2 – (a + x)^2
\]
Apply the difference of two squares identity:
\[
= (1 – (a + x))(1 + a + x)
\]
\[
= (1 – a – x)(1 + a + x)
\]
\(\textbf{Correct Answer: A} \)
Question 7
Which of the choices below is a factor of the expression:
\[
(a – b + c)^2 – (a + b – c)^2
\]
\[
\text{A) } a \quad
\text{B) } b \quad
\text{C) } c \quad
\text{D) } a-b \quad
\text{E } b+c
\]
Solution:
Using the difference of squares identity \( X^2 – Y^2 = (X – Y)(X + Y) \):
\[
= ((a – b + c) – (a + b – c))((a – b + c) + (a + b – c))
\]
Simplify inside the individual factors:
\[
= (a – b + c – a – b + c) (2a)
\]
\[
= (2c – 2b) \cdot 2a = 4a(c – b)
\]
Thus, \( a \) is a valid factor.
\(\textbf{Correct Answer: A} \)
Question 8
Identify a factor of the polynomial expression below:
\[
x^5 – 9x^3 + 8x^2 – 72
\]
\[
\text{A) } x-1 \quad
\text{B) } x+1 \quad
\text{C) } x-2 \quad
\text{D) } x+2 \quad
\text{E } x-4
\]
Solution:
Factor the expression by grouping the terms pairwise:
\[
x^5 – 9x^3 + 8x^2 – 72 = x^3(x^2 – 9) + 8(x^2 – 9)
\]
\[
= (x^2 – 9) (x^3 + 8)
\]
Decompose using the difference of squares and sum of cubes identities:
\[
= (x – 3)(x + 3)(x + 2)(x^2 – 2x + 4)
\]
Thus, \( x + 2 \) is a factor.
\(\textbf{Correct Answer: D} \)
Question 9
Which of the following is a factor of the expression:
\[
a^3 + 3a^2 + 3a + 2
\]
\[
\text{A) } a \quad
\text{B) } a+1 \quad
\text{C) } a+2 \quad
\text{D) } a^2+1 \quad
\text{E } a^2+2
\]
Solution:
Rewrite the constant term to isolate a perfect cube binomial expansion:
\[
a^3 + 3a^2 + 3a + 2 = a^3 + 3a^2 + 3a + 1 + 1
\]
\[
= (a + 1)^3 + 1^3
\]
Apply the sum of two cubes identity:
\[
= ((a + 1) + 1)((a + 1)^2 – (a + 1) + 1)
\]
\[
= (a + 2)(a^2 + a + 1)
\]
Thus, \( a + 2 \) is a factor.
\(\textbf{Correct Answer: C} \)
Question 10
Find a factor of the given expression:
\[
x^3 + x^2y – xy^2 – y^3 + (x + y)^2
\]
\[
\text{A) } x+y+1 \quad
\text{B) } x-y \quad
\text{C) } x-y+1 \quad
\text{D) } x \quad
\text{E } y
\]
Solution:
Rearrange and group the polynomial terms to factor out common binomial expressions:
\[
x^3 + x^2y – xy^2 – y^3 + (x + y)^2
\]
\[
= (x^3 – y^3) + xy(x – y) + (x + y)^2
\]
\[
= (x – y)(x^2 + xy + y^2) + xy (x – y) + (x + y)^2
\]
\[
= (x – y) (x^2 + xy + y^2 + xy) + (x + y)^2
\]
\[
= (x – y) (x + y)^2 + (x + y)^2
\]
Factor out the common term \( (x + y)^2 \):
\[
= (x + y)^2 (x – y + 1)
\]
Thus, \( x – y + 1 \) is a factor.
\(\textbf{Correct Answer: C} \)
QUESTION 11
If
\[
x – \frac{1}{x} = 2
\]
evaluate the following numerical value:
\[
x^3 – \frac{1}{x^3}
\]
\[
\text{A) } 10 \quad
\text{B) } 12 \quad
\text{C) } 14 \quad
\text{D) } 16 \quad
\text{E } 18
\]
Solution:
Cube both sides of the given equation:
\[
x \;- \; \frac{1}{x} = 2 \Rightarrow \left( x\; – \;\frac{1}{x} \right)^3 = 2^3
\]
\[
\Rightarrow x^3\; – \;3x^2 \cdot \frac{1}{x} + 3x \cdot \frac{1}{x^2} \;-\; \frac{1}{x^3} = 8
\]
\[
\Rightarrow x^3 \;-\; \frac{1}{x^3} \;- \;3\left(x \;-\; \frac{1}{x} \right) = 8
\]
Substitute the original value \( x – \frac{1}{x} = 2 \) into the expression:
\[
\Rightarrow x^3\; – \;\frac{1}{x^3}\; -\; 3 \cdot 2 = 8
\]
\[
\Rightarrow x^3\; – \;\frac{1}{x^3} = 14
\]
\(\textbf{Correct Answer: C} \)
Question 12
Given that
\[
\sqrt{x} + \frac{2}{\sqrt{x}} = 4
\]
determine the positive value of the following expression:
\[
\sqrt{x} – \frac{2}{\sqrt{x}}
\]
\[
\text{A) } \sqrt{ 2} \quad
\text{B) } 2\sqrt{ 2} \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E } 4\sqrt{ 2}
\]
Solution:
Let \( A = \sqrt{x} – \frac{2}{\sqrt{x}} \). Squaring both sides yields:
\[
A^2 = x + \frac{4}{x} – 4
\]
Next, square both sides of the given equation to find the value of the shared terms:
\[
\sqrt{x} + \frac{2}{\sqrt{x}} = 4 \Rightarrow \left( \sqrt{x} + \frac{2}{\sqrt{x}} \right)^2 = 4^2
\]
\[
\Rightarrow x + \frac{4}{x} + 4 = 16 \Rightarrow x + \frac{4}{x} = 12
\]
Substitute this value into our expression for \( A^2 \):
\[
A^2 = 12 – 4 = 8
\]
\[
\Rightarrow A = \sqrt{x} – \frac{2}{\sqrt{x}} = \pm \sqrt{8} = \pm 2\sqrt{2}
\]
Since we need the positive value, the result is \( 2\sqrt{2} \).
\(\textbf{Correct Answer: B} \)
QUESTION 13
Given the algebraic system:
\[
a + b + c = 6, \quad a^2 + b^2 + c^2 = 14, \quad \text{and} \quad \frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 33
\]
calculate the product:
\[
a \cdot b \cdot c
\]
\[
\text{A) } \frac{1}{3} \quad
\text{B) } 3 \quad
\text{C) } \frac{1}{2} \quad
\text{D) } 2 \quad
\text{E } 4\sqrt{ 2}
\]
Solution:
Square the first identity:
\[
a + b + c = 6 \Rightarrow (a + b + c)^2 = 6^2
\]
\[
\Rightarrow a^2 + b^2 + c^2 + 2(ab + ac + bc) = 36
\]
Substitute the known sum of squares (\( a^2+b^2+c^2 = 14 \)):
\[
\Rightarrow 14 + 2(ab + ac + bc) = 36 \Rightarrow ab + ac + bc = 11
\]
Find a common denominator for the fractional identity:
\[
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 33 \Rightarrow \frac{bc + ac + ab}{abc} = 33
\]
Substitute \( ab + ac + bc = 11 \) into the numerator:
\[
\frac{11}{abc} = 33 \Rightarrow abc = \frac{11}{33} = \frac{1}{3}
\]
\(\textbf{Correct Answer: A} \)
QUESTION 14
If \( \sqrt[3]{a} – \sqrt[3]{b} = 1 \) and \( a – b = 37 \), find the product value of \( a \cdot b \):
\[
\text{A) } 2^9 \quad
\text{B) } 3^9 \quad
\text{C) } 10^3 \quad
\text{D) } (11)^3 \quad
\text{E } (12)^3
\]
Solution:
Cube the given binomial radical difference equation:
\[
\sqrt[3]{a} \;- \;\sqrt[3]{b} = 1 \Rightarrow (\sqrt[3]{a} \;- \;\sqrt[3]{b})^3 = 1^3
\]
\[
\Rightarrow a \;- \;3\sqrt[3]{a^2}\sqrt[3]{b} + 3\sqrt[3]{a}\sqrt[3]{b^2}\; -\; b = 1
\]
\[
\Rightarrow (a – b) \;- \;3\sqrt[3]{ab}(\sqrt[3]{a}\; -\; \sqrt[3]{b}) = 1
\]
Substitute the known quantities \( a – b = 37 \) and \( \sqrt[3]{a} – \sqrt[3]{b} = 1 \):
\[
\Rightarrow 37\; – \;3\sqrt[3]{ab} \cdot (1) = 1
\]
\[
\Rightarrow 3\sqrt[3]{ab} = 36 \Rightarrow \sqrt[3]{ab} = 12
\]
Cube both sides to find the product:
\[
\Rightarrow ab = (12)^3
\]
\(\textbf{Correct Answer: E} \)
Question 15
Find the minimum possible value of the multivariable function:
\[
A = 2x^2 + y^2 – 2xy + 4x + 7
\]
\[
\text{A) } 1 \quad
\text{B) } 2 \quad
\text{C) } 3 \quad
\text{D) } 4 \quad
\text{E } 5
\]
Solution:
Decompose the polynomial terms to form complete squares:
\[
A = 2x^2 + y^2 – 2xy + 4x + 7
\]
\[
= (x^2 – 2xy + y^2) + (x^2 + 4x + 4) + 3
\]
\[
= (x – y)^2 + (x + 2)^2 + 3
\]
Since squared real expressions are non-negative (\( (x – y)^2 \geq 0 \) and \( (x + 2)^2 \geq 0 \)), the expression reaches its minimum value when both perfect squares equal zero:
\[
\underbrace{(x – y)^2}_{0} + \underbrace{(x + 2)^2}_{0} + 3 = 3
\]
\(\textbf{Correct Answer: C} \)
QUESTION 16
Given that \( x(x – 1) = a – 1 \) and \( x + 1 = b \), express \( x^3 \) in terms of variables \( a \) and \( b \):
\[
\text{A) } ab \quad
\text{B) } ab-1 \quad
\text{C) } ab+1 \quad
\text{D) } a+b-1 \quad
\text{E } a+b+1
\]
Solution:
Expand and rearrange the first equation:
\[
x(x – 1) = a – 1 \Rightarrow x^2 – x + 1 = a
\]
We are also given:
\[
x + 1 = b
\]
Multiply these two equations together to match the sum of cubes identity structure:
\[
(x + 1)(x^2 – x + 1) = ab
\]
\[
\Rightarrow x^3 + 1 = ab \Rightarrow x^3 = ab – 1
\]
\(\textbf{Correct Answer: B} \)
Question 17
If \( A = 2^{10} + 3^{10} + 2^6 \cdot 3^5 \), calculate the exact value of \( \sqrt{A} \):
\[
\text{A) } 271 \quad
\text{B) } 272 \quad
\text{C) } 273 \quad
\text{D) } 274 \quad
\text{E } 275
\]
Solution:
Structure the expression to look like a perfect square trinomial expansion \( (X+Y)^2 = X^2 + 2XY + Y^2 \):
\[
A = 2^{10} + 3^{10} + 2^6 \cdot 3^5 = (2^5)^2 + (3^5)^2 + 2 \cdot 2^5 \cdot 3^5
\]
\[
\Rightarrow A = (2^5 + 3^5)^2
\]
Taking the square root:
\[
\Rightarrow \sqrt{A} = 2^5 + 3^5 = 32 + 243 = 275
\]
\(\textbf{Correct Answer: E} \)
Question 18
Given that \( x^2 + x – 1 = 0 \), which of the following expressions is equal to \( x^5 \)?
\[
\text{A) } 5x-3 \quad
\text{B) } 5x+3 \quad
\text{C) } 3x-5 \quad
\text{D) } 3x+5 \quad
\text{E } x
\]
Solution:
Isolate the highest power term from the given equation:
\[
x^2 + x – 1 = 0 \Rightarrow x^2 = 1 – x
\]
Express \( x^5 \) in terms of lower power components using properties of exponents:
\[
x^5 = x \cdot (x^2)^2
\]
Substitute \( x^2 = 1 – x \):
\[
= x \cdot (1 – x)^2
\]
\[
= x \cdot (1 – 2x + x^2)
\]
Substitute \( x^2 = 1 – x \) again into the expression:
\[
= x \cdot (1 – 2x + 1 – x) = x \cdot (2 – 3x)
\]
\[
= 2x – 3x^2
\]
Perform one final substitution for \( x^2 \):
\[
= 2x – 3(1 – x) = 2x – 3 + 3x = 5x – 3
\]
\(\textbf{Correct Answer: A} \)
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