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(3(k+1)(k+2)+k(k+1)(k+2))/3=0
We multiply parentheses ..
(3(+k^2+2k+k+2)+k(+k^2+2k+k+2))/3=0
We multiply all the terms by the denominator
(3(+k^2+2k+k+2)+k(+k^2+2k+k+2))=0
We calculate terms in parentheses: +(3(+k^2+2k+k+2)+k(+k^2+2k+k+2)), so:
3(+k^2+2k+k+2)+k(+k^2+2k+k+2)
We multiply parentheses
3k^2+k^3+2k^2+k^2+6k+3k+2k+6
We do not support ekpression: k^3
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