Answer to Question #10686 in Algebra for nksprasad

Question

factorize a^3+b^3-c^3+3abc

Answer 1

factorize a³+b³-c³+3abc

(a + b + c)(a² + b² + c² + x)

a³ + ab² + ac² + ax
ba² + b³ + c²b + bx
ca² + cb² + c³ + cx

a³, b³, c³ are all valid
ab² + ac² + ba² + c²b + ca² + cb² + 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 + ab² + c²a +a( - ab - bc - ac) +
+ ba² + b³ + bc² + b( - ab - bc - ac) +
+ ca² + cb² + c³ + c(- ab - bc - ac) =
= a³ + b³ + c³ - 3abc

(a + b + c)(a² + b² + c² - ab - bc - ac) =
= a³ + b²a + c²a - a²b - abc - a²c +
+ ba² + b³ + c²b - ab² - b²c - abc +
+ ca² + cb² + c³ - acb -c²b - ac² =
= a³ + b³ + c³ - 3abc.

So,

a³+b³-c³+3abc = (a + b + c)(a² + b² + c² - ab - bc - ac).