Question : What is the value of $\left(27x^3 + 58x^2 y + 31xy^2 + 8y^3\right)$, when $x = 5$ and $y = –7$?
Option 1: 1924
Option 2: –1926
Option 3: –1924
Option 4: 1926
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Correct Answer: –1924
Solution :
Given: $x = 5 $ and $y = -7$
$\left(27x^3 + 58x^2y + 31xy^2 + 8y^3\right)$
= $27x^3+54x^2y+36xy^2+8y^3+ 4x^{2}y - 5xy^{2}$
= $(3x + 2y)^{3} + 4x^{2}y - 5xy^{2}$
= $(3 \times 5 + 2 \times (-7)^{3} + 4 \times 5^{2} \times (-7) -(5\times 5 \times (-7)^{2})$
= $(1)^{3}-700-1225$
= $-1924$
Hence, the correct answer is –1924.
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