subtract 3a (a+d+c)+2b(a_b+c)from 4c (_a+b+c)
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First we will add 3a(a+d+c)and 2b(a-b+c) is 3a2+3ad+3ac and 2ba-2b2+2bc, then we will subtract 3a2+3ad+3ac+2ba-2b2+2bc from - 4ca+4cb+4c2 so the answer is 4ca+4cb+4c2-3a2-3ad-3ac-2ba+2b2-2bc
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