Question

x3+y3-12xy+64 when x+y= 4​

Answer 1

hey !!

Here's your answer !!

we can solve this question by using this identity

x³+y³+z³-3xyz=(x+y+z)(x²+y²+z²-xy-yz-zx)

now

(x)³+(y)³+(4)³-3*x*y*4= (x+y+4)[(x)²+(y)²+(4)²-xy-4y-4x]

x³+y³+64-12xy=(4+4)(x²+y²+16-xy-4y-4x)

x³+y³+64-12xy=8x²+8y²+128-8xy-32y-32x ( required answer )

hope it helps !!

Answer 2

taking cube both sides we get

(x+y)^3=4^3=64

now expanding x^3+ y^3-3xy(x-y)=64

putting x-y=4 we get

x^3+y^3-3xy(4)=64

or x^3+y^3-12xy+64=0