1=(3b*3b*3b)+22-45

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Solution for 1=(3b*3b*3b)+22-45 equation: 



1=(3b*3b*3b)+22-45

We move all terms to the left:

1-((3b*3b*3b)+22-45)=0

We add all the numbers together, and all the variables

-((+3b*3b*3b)+22-45)+1=0



We calculate terms in parentheses: -((+3b*3b*3b)+22-45), so:

(+3b*3b*3b)+22-45

We add all the numbers together, and all the variables

(+3b*3b*3b)-23

We get rid of parentheses

3b*3b*3b-23

Wy multiply elements

27b^3*3-23

We do not support ebpression: b^3

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