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

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

Factor

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

Solution Steps

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

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

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 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).

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

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

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