Question

Given that [x2 + y2 = z2], then find the value of [x6 + y6-z6]/[x2y2z2].

Answer 1

Answer:

= > x^2 + y^2 = z^2

Cube on both sides:

= > ( x^2 + y^2 )^3 = ( z^2 )^3

Using ( a + b )^3 = a^3 + b^3 + 3ab( a + b )

= > x^6 + y^6 + 3$x^2 y^2 (x^2+y^2)$ = z^6

x^2 + y^2 = z^2

= > x^6 + y^6 + 3( x^2 y^2 )( z^2 ) = z^6

= > x^6 + y^2 + 3x^2 y^2 z^2 = z^6

= > x^6 + y^6 - z^6 = - 3x^2 y^2 z^2

= > ( x^2 + y^2 - z^6 ) / ( x^2 y^2 z^2 ) = - 3

Required value is - 3.