Factoring Differences of Cubes
Practice Problems

Since $$\blue a$$ is the cube root of the first term, $$\blue a = \sqrt[3]{x^3} = \blue x$$.

Similarly, since $$ \red b$$ is the cube root of the second term, $$\red b = \sqrt[3] 8 = \red 2$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red { b } + \red{ b}^2)\\ x^3 - 8 & = (\blue x - \red 2)(\blue x^2 + \blue x\cdot \red { 2 } + \red { 2}^2)\\ & = (x - 2)(x^2 + 2x + 4) \end{align*} $$

$$ x^3 - 8 = (x - 2)(x^2 + 2x +4) $$

Since $$\blue a$$ is the cube root of the first term, $$\blue a = \sqrt[3]{x^3} = \blue x$$.

Likewise, since $$\red b$$ is the cube root of the second term, $$\red b = \sqrt[3] 1 = \red 1$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ x^3 - 1 & = (\blue x - \red 1)(\blue x^2 + \blue x \cdot \red 1 + \red 1^2)\\ & = (x - 1)(x^2 + x + 1) \end{align*} $$

$$ x^3 - 1 = (x - 1)(x^2 + x + 1) $$

Since $$\blue a$$ is the cube root of the first term, $$\blue a = \sqrt[3]{27x^3} = \blue{3x }$$.

Likewise, since $$\red b$$ is the cube root of the second term, $$\red b = \sqrt[3]{64} = \red 4$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 27x^3 - 64 & = (\blue{3x} - \red 4)[\blue{(3x)}^2 + \blue{(3x)}\red{(4)} + \red 4^2]\\ & = (3x - 4)(9x^2 + 12x + 16) \end{align*} $$

$$ 27x^3 - 64 = (3x - 4)(9x^2 + 12x + 16) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{8x^3} = 2x$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{125} = 5$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 8x^3 - 125 & = (\blue{2x} - \red 5)[\blue{(2x)}^2 + \blue{(2x)}\red{(5)} + \red 5^2]\\ & = (2x - 5)(4x^2 + 10x + 25) \end{align*} $$

$$ 8x^3 - 125 = (2x - 5)(4x^2 + 10x + 25) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{x^3} = x$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{y^3} = y$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ x^3 - y^3 & = (\blue x - \red y)(\blue x^2 + \blue x \red y + \red y^2) \end{align*} $$

$$ x^3 - y^3 = (x - y)(x^2 + xy + y^2) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{216x^3} = 6x$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{27y^3} = 3y$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 216x^3 - 27y^3 & = (\blue{6x} - \red{3y})[\blue{(6x)}^2 + \blue{(6x)}\red{(3y)} + \red{(3y)}^2]\\ & = (6x - 3y)(36x^2 + 18xy + 9y^2)\\ & = 3(2x - y)\cdot 9(4x^2 + 2xy + y^2)\\ &= 27(2x - y)(4x^2 + 2xy + y^2)\\ \end{align*} $$

$$ 216x^3 - 27y^3 = 27(2x - y)(4x^2 + 2xy + y^2) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{8x^6} = 2x^2$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{125y^9} = 5y^3$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 8x^6 - 125y^9 & = (\blue{2x^2} - \red{5y^3})[\blue{(2x^2)}^2 + \blue{(2x^2)}\red{(5y^3)} + \red{(5y^3)}^2]\\ & = (2x^2 - 5y^3)(4x^4 + 10x^2y^3 + 25y^6) \end{align*} $$

$$ 8x^6 - 125y^9 = (2x^2 - 35^3)(4x^4 + 10x^2y^3 + 25y^6) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{64x^{3/2}} = (64x^{3/2})^{1/3} = 4x^{1/2}$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{343y^{6/5}} = (343y^6)^{1/3} = 7y^2$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 64x^{3/2} - 343y^6 & = \left(\blue{4x^{1/2}} - \red{7y^2}\right)\left[\blue{\left(4x^{1/2}\right)}^2 + \blue{\left(4x^{1/2}\right)}\red{\left(7y^2\right)} + \red{\left(7y^2\right)}^2\right]\\ & = \left(4x^{1/2} - 7y^2\right)\left(16x + 28x^{1/2}y^2 + 49y^4\right) \end{align*} $$

$$ 64x^{3/2} - 343y^6 = \left(4x^{1/2} - 7y^2\right)\left(16x + 28x^{1/2}y^2 + 49y^4\right) $$

$$ 5x^{12} - 135y^{30} = 5\left(x^{12} - 27y^{30}\right) $$

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{x^{12}} =x^4$$.

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{27y^{30}} = 3y^{10}$$.

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 5\left(x^{12} - 27y^{30}\right) & = 5\left(\blue{x^4} - \red{3y^{10}}\right)\left[\blue{\left(x^4\right)}^2 - \blue{\left(x^4\right)}\red{\left(3y^{10}\right)} + \red{\left(3y^{10}\right)}^2\right]\\ & = 5\left(x^4 - 3y^{10}\right)\left(x^8 - 3x^4y^{10} + 9y^{20}\right)\\ \end{align*} $$

$$ 5x^{12} - 135y^{30} = 5\left(x^4 - 3y^{10}\right)\left(x^8 - 3x^4y^{10} + 9y^{20}\right) $$

$$ 24x^{21} - 375y^{15} = 3\left(8x^{21} - 125y^{15}\right) $$

Since $$a$$ is the cube root of the first term.

$$ a = \sqrt[3]{8x^{21}} = 2x^7 $$

Likewise, since $$b$$ is the cube root of the second term,

$$ b = \sqrt[3]{125y^{15}} = 5y^5 $$

$$ \begin{align*} a^3 - b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ 3\left(8x^{21} - 125y^{15}\right) & = 3\left(\blue{2x^7} - \red{5y^5}\right)\left[\blue{\left(2x^7\right)}^2 + \blue{\left(2x^7\right)}\red{\left(5y^5\right)} + \blue{\left(5y^5\right)}^2\right]\\ & = 3\left(2x^7 - 5y^5\right)\left(4x^{14} + 10x^7y^5 + 25y^{10}\right)\\ \end{align*} $$

$$ 24x^{21} - 375y^{15} = 3\left(2x^7 - 5y^5\right)\left(4x^{14} + 10x^7y^5 + 25y^{10}\right) $$