If $x + y + 4 = 0,$ find the value of $x^{3} + y^{3} - 12 x y + 64 .$

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Solution

Finding the value of the given expression:
Let $a = x, b = y a n d c = 4$
$a + b + c = x + y + 4 = 0 (G i v e n)$
Using identity: $a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) (a^{2} + b^{2} + c^{2} - a b - b c - c a)$
So, if $a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 a b c$
$x^{3} + y^{3} + {(4)}^{3} = 3 \times 4 x y x^{3} + y^{3} + 64 = 12 x y x^{3} + y^{3} - 12 x y + 64 = 0$
Hence, the value of $x^{3} + y^{3} - 12 x y + 64$ is $0 .$

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Find the value of:
(i) $x^{3} + y^{3} - 12 x y + 64$ , when x + y = - 4

(ii) $x^{3} - 8 y^{3} - 36 x y - 216$ , when x = 2y + 6

x^{3} + y^{3} - 12 x y + 64,

x + y = - 4.

if x +y+ 4 =0 , then find the value of x³ + b³- 12xy + 64

x^{3} + y^{3} - 12 x y + 64

x + y = 4

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