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Question #10686
factorize a^3+b^3-c^3+3abc
factorize a&sup3;+b&sup3;-c&sup3;+3abc

(a + b + c)(a&sup2; + b&sup2; + c&sup2; + x)

a&sup3; + ab&sup2; + ac&sup2; + ax
ba&sup2; + b&sup3; + c&sup2;b + bx
ca&sup2; + cb&sup2; + c&sup3; + cx

a&sup3;, b&sup3;, c&sup3; are all valid
ab&sup2; + ac&sup2; + ba&sup2; + c&sup2;b + ca&sup2; + cb&sup2; + ax + bx + cx are not.

They need to cancel out and any variable multiplied on the other side would give you the square of that variable

- ab - bc - ac

This really does check out:

= a^3 + ab&sup2; + c&sup2;a +a( - ab - bc - ac) +
+ ba&sup2; + b&sup3; + bc&sup2; + b( - ab - bc - ac) +
+ ca&sup2; + cb&sup2; + c&sup3; + c(- ab - bc - ac) =
= a&sup3; + b&sup3; + c&sup3; - 3abc

(a + b + c)(a&sup2; + b&sup2; + c&sup2; - ab - bc - ac) =
= a&sup3; + b&sup2;a + c&sup2;a - a&sup2;b - abc - a&sup2;c +
+ ba&sup2; + b&sup3; + c&sup2;b - ab&sup2; - b&sup2;c - abc +
+ ca&sup2; + cb&sup2; + c&sup3; - acb -c&sup2;b - ac&sup2; =
= a&sup3; + b&sup3; + c&sup3; - 3abc.

So,

a&sup3;+b&sup3;-c&sup3;+3abc = (a + b + c)(a&sup2; + b&sup2; + c&sup2; - ab - bc - ac).

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