Factorise the following:
(x/2+y+z/3)^3+[x/3-2y/3+z]^3+ - 5x/6-y/3- 4z/3
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Your question is typo mistake. I guess (x/2+y+z/3)3+(x/3-2y/3+z)3+(-5x/6+y/3-4z/3)3 should be (x/2+y+z/3)3+(x/3-2y/3+z)3+(-5x/6-y/3-4z/3)3. Like this, a+b+c=0 would be true
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