If $x + y + 4 = 0$ , find the value of $x^{2} + y^{2} - 12 x y + 64$

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Solution

It is given that $x + y + 4 = 0$ or $x + y = - 4$

We know the identity $(a + b)^{3} = a^{3} + b^{3} + 3 a b (a + b)$

Now consider $x + y = - 4$ and take cube on both sides and then appy the above identity as follows:

$(x + y)^{3} = (- 4)^{3} \Rightarrow x^{3} + y^{3} + 3 x y (x + y) = - 64 \Rightarrow x^{3} + y^{3} + 3 x y (- 4) = - 64 \Rightarrow x^{3} + y^{3} - 12 x y + 64 = 0$

Hence, $x^{3} + y^{3} - 12 x y + 64 = 0$

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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

y = \frac{1}{2} {sin}^{- 1} (\frac{2 x y}{x^{2} + y^{2}})

lim x y \to 0 =

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