what is the tricky way for solving the question of class 9th ncert math of Polynomials chapter of ncert of exercise 2.5 of math give me the best and simple way for solving this question in easy and simple way Factorise: 27×3+y3+z3–9xyz
Share
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
The expression27x3+y3+z3–9xyz can be written as (3x)3+y3+z3–3(3x)(y)(z)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
We know that, x3+y3+z3–3xyz = (x+y+z)(x2+y2+z2–xy –yz–zx)
27x3+y3+z3–9xyz = (3x)3+y3+z3–3(3x)(y)(z)
= (3x+y+z)[(3x)2+y2+z2–3xy–yz–3xz]
= (3x+y+z)(9x2+y2+z2–3xy–yz–3xz)