Question

factorise : 125x^3 - 64y^3.​

Answer 1

$\huge\boxed{\fcolorbox{red}{grey}{Solution}}$

${125x}^{3} - {64y}^{3} \\ {5x}^{3} - {4y}^{3} \\ (5x - 4y)( {5x}^{2} + {4y}^{2} - 20xy) \\ (5x - 4y)( {25x}^{2} + {16y}^{2} - 20xy)$

Answer 2

HERE IS YOUR ANSWER

125X^3—64Y^3=0

(5X—4Y)(25X^2+20XY+16Y^2)=0

SO WE GET,

5X—4Y=0

SO,

5X=4Y

BY OTHER YOU WILL GET IMAGINARY ROOTS OR PRECISE IMAGINARY RELATION

HENCE DONE

CHEERS

MARK BRAINLIEST ⭐