Question

if x - y = 1 ,X Y =28 find x cube minus y cube​

Answer 1

ANSWER

$\boxed{\textbf{\large{xcube-ycube=85}}}$

Given

x - y = 1

xy = 28

To find :

x^3-y^3 =?

Explanation

we have given

x - y = 1 , x y = 28

we know the algebraic formula,

(a^3 - b^3 )= ( a - b) (a^2 + ab + b^2)

therefor,

(x^3 - y^3) = ( x - y) (x ^2 + xy + y^2)

we have,

(x - y) = 1

(xy) = 28

(x^2 + y ^2) = ?

we know,

(a - b) ^2 = a^2 - 2ab + b^2

therefor,

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

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

therefor

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

= (1)^2 + 2(28)

= 1 + 56

= 57

so,

(x^3 - y^3) = ( x - y) (x ^2 + xy + y^2)

= (1) (( x^2 + y^2) + ( xy) )

= (1) ((57)+( 28))

= (1) (85)

= [85]

finally,

$\boxed{\textbf{\large{xcube-ycube=85}}}$