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(x+1)^x2+7x+10=(2x-3)x2+7+10
We move all terms to the left:
(x+1)^x2+7x+10-((2x-3)x2+7+10)=0
We add all the numbers together, and all the variables
7x+(x+1)^x2-((2x-3)x2+7+10)+10=0
We calculate terms in parentheses: -((2x-3)x2+7+10), so:
(2x-3)x2+7+10
We add all the numbers together, and all the variables
(2x-3)x2+17
We multiply parentheses
2x^3-3x^2+17
We do not support expression: x^3
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