factorise the following equation i will mark your ans as branliest.
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(x−y−x)(x2+y2+z2+xy−yz+xz)
According to the equation,
a3−b3−c3−3abc=(a−b−c)(a2+b2+c2+ab−bc+ca)
The given question can be written as
=(x3+xy2+x2y−xyz+x2z−x2y−y3−yz2−xy2+y2z−xyz−x2z−y2z−z3−xyz+yz2−xz2)
So we get,
=(x3−y3−z3−xyz−xyz−xyz)
=x3−y3−z3−3xyz
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