Solve 7x^10+56x^7y^6

$$7 x ^ { 10 } + 56 x ^ { 7 } y ^ { 6 }$$

Factor

$$7\left(x+2y^{2}\right)\left(x^{2}-2xy^{2}+4y^{4}\right)x^{7}$$

Solution Steps

Factor out 7.

$$7\left(x^{10}+8x^{7}y^{6}\right)$$

Consider x^{10}+8x^{7}y^{6}. Factor out x^{7}.

$$x^{7}\left(x^{3}+8y^{6}\right)$$

Consider x^{3}+8y^{6}. Rewrite x^{3}+8y^{6} as x^{3}+\left(2y^{2}\right)^{3}. The sum of cubes can be factored using the rule: a^{3}+b^{3}=\left(a+b\right)\left(a^{2}-ab+b^{2}\right).

$$\left(x+2y^{2}\right)\left(x^{2}-2xy^{2}+4y^{4}\right)$$

Rewrite the complete factored expression.

$$7x^{7}\left(x+2y^{2}\right)\left(x^{2}-2xy^{2}+4y^{4}\right)$$

Evaluate

$$7\left(x^{3}+8y^{6}\right)x^{7}$$