How to solve (x)^(7)-x(y)^(6)

1. Factor the equation (x + 2)(x + 3) = 0. 2. Set each factor equal to zero: x + 2 = 0 and x + 3 = 0. 3. Solve for x: x = -2 and x = -3.

Solve (x)^(7)-x(y)^(6) | Equatix

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Question

$${ x }^{ 7 } -x { y }^{ 6 }$$

Factor

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

Solution Steps

Factor out $x$.

$$x\left(x^{6}-y^{6}\right)$$

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

$$\left(x^{3}-y^{3}\right)\left(x^{3}+y^{3}\right)$$

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

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

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

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

Rewrite the complete factored expression.

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

Show Solution

Evaluate

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