Question : What is the value of $64x^{3} + 36x^{2}y + 24xy^{2} + 2y^{3}$, when $x = 3$ and $y = -4?$
Option 1: $304$
Option 2: $1456$
Option 3: $1584$
Option 4: $432$
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Correct Answer: $1456$
Solution :
Given: $x = 3, y = -4$
So, the value of $64x^{3} + 36x^{2}y + 24xy^{2} + 2y^{3}$
$=x^{2}(64x + 36y) + y^{2}(24x + 2y)$
$=3^{2}[(64 × 3) + (36 × -4)] + (-4)^{2}[(24 × 3) + (2 × -4)]$
$=9 × [192 -144] + 16 × [72 -8]$
$=(9 × 48) + (16 × 64)$
$=432 + 1024$
$=1456$
Hence, the correct answer is $1456$.
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