Find $x^{3} + y^{3} - 12 x y + 64$ , when $x + y = 4$ .

Open in App

Solution

$(x + y)^{2} = x^{2} + y^{2} + 2 x y$
$4^{2} = x^{2} + y^{2} + 2 x y$
$16 - 2 x y = x^{2} + y^{2}$
$x^{2} + y^{2} = 16 - 2 x y$
now,
$x^{3} + y^{3} = (x + y) (x^{2} + y^{2} - x y)$
$x^{3} + y^{3} = (4) (16 - 2 x y - x y)$
$x^{3} + y^{3} = (4) (16 - 3 x y)$
$x^{3} + y^{3} = 64 - 12 x y$
$x^{3} + y^{3} - 64 + 12 x y = 0$

Suggest Corrections

Similar questions

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

x + y = - 4.

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

Join BYJU'S Learning Program