Question

⇒Find the value of x³ + y³ + 12xy - 64 when x + y = 4.

Answer 1

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if x + y = 4, x^3 + y^3 + 12xy - 64 = ?
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use identity : a^3 + b^3 + c^3 - 3abc

= ( a + b + c ) (a^2 + b^2 + c^2 - ab - bc - ca )
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= x^3 + y ^3 - 3( x )( y )(-4 ) + ( - 4 )^3

= x^3 + y^3 + ( - 4) ^3 - 3 ( x ) ( y ) (-4)

= ( x + y - 4 ) (x^2 + y^2 + (-4 )^2 -xy - y(-4) - (-4 )(x )

since , x + y = 4 , what's now ?

= ( 4 - 4 ) ( x^2 + y^2 + 16 - xy + 4y + 4 x )

= 0 × ( x^2 + y^2 + 16 - xy + 4y + 4x )

= 0

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Your Answer = 0
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