Question

factorise : ab(a²+b²-c²) -bc(c²-a²-b²)+ca(a²+b²-c²)​

Answer 1

Answer:

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Answer 2

Tips for factorization

1. First, group out the common factors.
2. Check whether you can apply some identity.

Note that $c^2-a^2-b^2=-(a^2+b^2-c^2)$.

Given to factorize:

$ab(a^2+b^2-c^2)+bc(a^2+b^2-c^2)+ca(a^2+b^2-c^2)$

$=(ab+bc+ca)(a^2+b^2-c^2)$

More information:

If you learned about the factor theorem, you can use it to factorize some higher degree polynomial.

If $\alpha$ is a zero, $(x-\underline{\alpha })$ is a factor.

$\alpha$ is what makes $f(\underline{\alpha })=0$.

Using the fact, we can factorize $x^2-10x+1$.

$5\pm2\sqrt{6}$ is what makes $x^2-10x+1=0$.

Then $x^2-10x+1=(x-5-2\sqrt{6} )(x-5+2\sqrt{6} )$.

Or we can use identity.

$x^2-10x+1=(x-5)^2-24$

$\implies x^2-10x+1=(x-5)^2-(2\sqrt{6} )^2$

$\implies x^2-10x+1=(x-5-2\sqrt{6} )(x-5+2\sqrt{6} )$