Solve 3sq^4-3sj^4

$$3 s q ^ { 4 } - 3 s j ^ { 4 }$$

Factor

$$3s\left(q-j\right)\left(q+j\right)\left(q^{2}+j^{2}\right)$$

Solution Steps

Factor out 3.

$$3\left(sq^{4}-sj^{4}\right)$$

Consider sq^{4}-sj^{4}. Factor out s.

$$s\left(q^{4}-j^{4}\right)$$

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

$$\left(q^{2}-j^{2}\right)\left(q^{2}+j^{2}\right)$$

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

$$\left(q-j\right)\left(q+j\right)$$

Rewrite the complete factored expression.

$$3s\left(q-j\right)\left(q+j\right)\left(q^{2}+j^{2}\right)$$

Evaluate

$$3s\left(q^{4}-j^{4}\right)$$