The Total solution for NCERT class 6-12
8x2y3┬атАУ x5┬а= x2(8y3┬атАУX3)= x2(2y)3┬атАУ (x)3]= x2[(2y тАУ x) (2y)2┬а+ 2y x x + (x)2]= x2(2y тАУ x) (4y2┬а+ 2xy + x2)