Solve 2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^2-a^4-b^4-c^4

$$2 a ^ { 2 } b ^ { 2 } + 2 a ^ { 2 } c ^ { 2 } + 2 b ^ { 2 } c ^ { 2 } - a ^ { 4 } - b ^ { 4 } - c ^ { 4 }$$

Factor

$$\left(a+b-c\right)\left(a-b+c\right)\left(c+b-a\right)\left(a+b+c\right)$$

Solution Steps

Rewrite 2a^{2}b^{2}+2a^{2}c^{2}+2b^{2}c^{2}-a^{4}-b^{4}-c^{4} as \left(2bc\right)^{2}-\left(-a^{2}+b^{2}+c^{2}\right)^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(2bc+a^{2}-b^{2}-c^{2}\right)\left(2bc-a^{2}+b^{2}+c^{2}\right)$$

Reorder the terms.

$$\left(a^{2}-b^{2}+2bc-c^{2}\right)\left(-a^{2}+b^{2}+2bc+c^{2}\right)$$

Consider a^{2}-b^{2}+2bc-c^{2}. Rewrite as difference of two terms.

$$a^{2}-\left(b^{2}-2bc+c^{2}\right)$$

Rewrite the terms as squares.

$$a^{2}-\left(b-c\right)^{2}$$

The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(a-b+c\right)\left(a+b-c\right)$$

Consider -a^{2}+b^{2}+2bc+c^{2}. Rewrite -a^{2}+b^{2}+2bc+c^{2} as \left(b+c\right)^{2}-a^{2}. The difference of squares can be factored using the rule: p^{2}-q^{2}=\left(p-q\right)\left(p+q\right).

$$\left(b+c-a\right)\left(b+c+a\right)$$

Reorder the terms.

$$\left(-a+b+c\right)\left(a+b+c\right)$$

Rewrite the complete factored expression.

$$\left(-a+b+c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)$$

Evaluate

$$\left(-\left(c-a\right)^{2}+b^{2}\right)\left(\left(a+c\right)^{2}-b^{2}\right)$$