Find the value of $x^{3} - 8 y^{3} - 36 x y - 216$ when $x = 2 y + 6$ .

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Solution

Given,

$x = 2 y + 6$
$=> x - 2 y - 6 = 0$ ...(i)
$x^{3} - 8 y^{3} - 36 x y - 216$
$= (x)^{3} + (- 2 y)^{3} + (- 6)^{3} - 3 \times (x) \times (- 2 y) \times (- 6)$
$= (x - 2 y - 6) (x^{2} + 4 y^{2} + 36 + 2 x y - 12 y + 6 x)$
$= 0 (x^{2} + 4 y^{2} + 36 + 2 x y - 12 y + 6 x)$ ...from (i)
$= 0$

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x = 2 y + 6

x^{3} - 8 y^{3} - 36 x y - 216

Question 39
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

Question 39
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

If $x - 2 y = 11$ and $x y = 8$ find the value of $x^{3} - 8 y^{3}$ .

The value of $(2 y - x)^{3}$ is _________.

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