Factoring Sums of Cubes - Practice Problems , explained step by step with fully downloadable pdf worksheet

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Problem 1

Factor $$x^3 + 8$$.

Step 1

Identify $$\blue a$$ and $$\red b$$.

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

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

Step 2

Write down the factored form.

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

Answer

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

Problem 2

Factor $$x^3 + 64$$.

Step 1

Identify $$\blue a$$ and $$\red b$$.

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

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

Step 2

Write down the factored form.

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

Answer

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

Problem 3

Factor $$8x^3 + 27$$.

Step 1

Identify $$\blue a$$ and $$ \red b $$ .

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

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

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 8x^3 + 27 & = (\blue{2x} + \red 3)[\blue{(2x)}^2 - \blue{(2x)}\red{(3)} + \red 3^2]\\ & = (2x + 3)(4x^2 - 6x + 9) \end{align*} $$

Answer

$$ 8x^3 + 27 = (2x + 3)(4x^2 - 6x + 9) $$

Problem 4

Factor $$64x^3 + 125$$.

Step 1

Identify $$a$$ and $$b$$.

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

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

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 64x^3 + 125 & = (\blue{4x} + \red 5)[\blue{(4x)}^2 - \blue{(4x)}\red{(5)} + \red 5^2]\\ & = (4x + 5)(16x^2 - 20x + 25) \end{align*} $$

Answer

$$ (4x + 5)(16x^2 - 20x + 25) $$

Problem 5

Factor $$x^3 + y^3$$.

Step 1

Identify $$a$$ and $$b$$.

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, $$b = \sqrt[3]{y^3} = y$$.

Step 2

Write down the factored form.

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

Answer

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

Problem 6

Factor $$216x^3 + 125y^3$$.

Step 1

Identify $$a$$ and $$b$$.

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

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 216x^3 + 125y^3 & = (\blue{6x} + \red{5y})[\blue{(6x)}^2 - \blue{(6x)}\red{(5y)} + \red{(5y)}^2]\\ & = (6x + 5y)(36x^2 - 30xy + 25y^2) \end{align*} $$

Answer

$$ 216x^3 + 125y^3 = (6x + 5y)(36x^2 - 30xy + 25y^2) $$

Sum Of Cubes Calculator

Problem 7

Factor $$8x^6 + 27y^9$$ as a sum of cubes.

Step 1

Identify $$a$$ and $$b$$.

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, $$b = \sqrt[3]{27y^9} = 3y^3$$.

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 8x^6 + 27y^9 & = (\blue{2x^2} + \red{3y^3})[\blue{(2x^2)}^2 - \blue{(2x^2)}\red{(3y^3)} + \red{(3y^3)}^2]\\ & = (2x^2 + 3y^3)(4x^4 - 6x^2y^3 + 9y^6) \end{align*} $$

Answer

$$ 8x^6 + 27y^9 = (2x^2 + 3y^3)(4x^2 - 6x^2y^3 + 9y^6) $$

Problem 8

Factor $$1000x^{3/2} + 343y^{6/5}$$ as a difference of cubes.

Step 1

Identify $$a$$ and $$b$$.

Since $$a$$ is the cube root of the first term, $$a = \sqrt[3]{1000x^{3/2}} = (1000x^{3/2})^{1/3} = 10x^{1/2}$$.

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

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 1000x^{3/2} + 343y^{6/5} & = \left(\blue{10x^{1/2}} + \red{7y^{2/5}}\right)\left[\blue{\left(10x^{1/2}\right)}^2 - \blue{\left(10x^{1/2}\right)}\red{\left(7y^{2/5}\right)} + \red{\left(7y^{2/5}\right)}^2\right]\\ & = \left(10x^{1/2} + 7y^{2/5}\right)\left(100x - 70x^{1/2}y^{2/5} + 49y^{4/5}\right) \end{align*} $$

Answer

$$ 1000x^{3/2} + 343y^{6/5} = \left(10x^{1/2} + 7y^{2/5}\right)\left(100x - 70x^{1/2}y^{2/5} + 49y^{4/5}\right) $$

Problem 9

Factor $$3x^2 + 7y^4$$ as a sum of cubes.

Step 1

Identify $$a$$ and $$b$$.

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

Likewise, since $$b$$ is the cube root of the second term, $$b = \sqrt[3]{7y^4}$$.

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 3x^2 + 7y^4 & = \left(\blue{\sqrt[3]{3x^2}} + \red{\sqrt[3]{7y^4}}\right)\left[\blue{\left(\sqrt[3]{3x^2}\right)}^2 - \blue{\left(\sqrt[3]{3x^2}\right)}\red{\left(\sqrt[3]{7y^4}\right)} + \blue{\left(\sqrt[3]{7y^4}\right)}^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left[\left(\sqrt[3] 3\,x^{2/3}\right)^2 - \sqrt[3]{3x^2\cdot 7y^4} + \left(\sqrt[3]7\,y^{4/3}\right)^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left[\left(\sqrt[3] 3\,x^{2/3}\right)^2 - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \left(\sqrt[3]7\,y^{4/3}\right)^2\right]\\ & = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left(\sqrt[3] 9\,x^{4/3} - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \sqrt[3]{49}\,y^{8/3}\right) \end{align*} $$

Answer

$$ 3x^2 + 7y^4 = \left(\sqrt[3] 3\,x^{2/3} + \sqrt[3] 7\,y^{4/3}\right)\left(\sqrt[3] 9\,x^{4/3} - \sqrt[3]{21}\,x^{2/3}y^{4/3} + \sqrt[3]{49}\,y^{8/3}\right) $$

Problem 10

Factor $$24x^{21} + 375y^{15}$$ as a sum of cubes.

Step 1

Identify $$a$$ and $$b$$.

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

$$ \begin{align*} a & = \sqrt[3]{24x^{21}}\\ & = \sqrt[3]{8\cdot 3 x^{21}}\\ & = 2\sqrt[3]{3 x^{21}}\\ & = 2\sqrt[3]3\,x^7 \end{align*} $$

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

$$ \begin{align*} b & = \sqrt[3]{375y^{15}}\\ & = \sqrt[3]{125\cdot 3y^{15}}\\ & = 5\sqrt[3] 3\, y^5 \end{align*} $$

Step 2

Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a + \red b)(\blue a^2 - \blue a \red b + \red b^2)\\ 24x^{21} + 375y^{15} & = \left(\blue{2\sqrt[3]3\,x^7} + \red{5\sqrt[3] 3\, y^5}\right)\left[\blue{\left(2\sqrt[3]3\,x^7\right)}^2 - \blue{\left(2\sqrt[3]3\,x^7\right)}\red{\left(5\sqrt[3] 3\, y^5\right)} + \blue{\left(5\sqrt[3] 3\, y^5\right)}^2\right]\\ & = \left(2\sqrt[3]3\,x^7 + 5\sqrt[3] 3\, y^5\right)\left(4\sqrt[3]9\,x^{14} - 10\sqrt[3]9\,x^7y^5 + 25\sqrt[3] 9\, y^{10}\right)\\ & = \sqrt[3] 3\left(2x^7 + 5y^5\right)\cdot \sqrt[3] 9 \left(4x^{14} - 10x^7y^5 + 25y^{10}\right)\\ & = \sqrt[3]{27}\left(2x^7 + 5y^5\right)\left(4x^{14} - 10x^7y^5 + 25y^{10}\right)\\ & = 3\left(2x^7 + 5y^5\right)\left(4x^{14} - 10x^7y^5 + 25y^{10}\right) \end{align*} $$

Answer

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

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