Question

using identities x-y=6, xy=4, find the value of x3-y3?

Answer 1

x³-y³=(x-y)(x²+y²+xy)=(x-y)[(x-y)²+2xy+xy]
=(x-y)[(x-y)²+3xy]
=6*[6²+3*4]
=6*[36+12]
=6*48
=288

Answer 2

Answer:

$x^3-y^3=288$

Step-by-step explanation:

Identity : $x^3-y^3=(x-y)(x^2+y^2+xy)$

it can be re-written as:

$x^3-y^3=(x-y)[(x-y)^2+2xy+xy]$

$x^3-y^3=(x-y)[(x-y)^2+3xy]$

Since we are given that  x-y=6, xy=4

$x^3-y^3=(6)[(6)^2+3(4)]$

$x^3-y^3=(6)[36+12]$

$x^3-y^3=(6)[48]$

$x^3-y^3=288$