The Total solution for NCERT class 6-12
We know that a3┬а+ b3┬а= (a + b)(a2┬атАУ ab + b2)a3┬атАУ b3┬а= (a тАУ b) (a2┬а+ aft + b2)p3┬а+ 21 = (p)3┬а+ (3)3= (p + 3) (p2тАУ p x 3 + 32)= (p + 3) (p2┬атАУ 3p + 9)