# Sum and Difference of Cubes

## Related calculator: Factoring Polynomials Calculator

Sum and Difference of Cubes:

$\color{purple}{a^3 \pm b^3=\left(a\pm b\right)\left(a^2 \mp ab+b^2 \right)}$

Proof of this fact is straightforward.

We just prove it from right to left.

${\left({a}-{b}\right)}{\left({{a}}^{{2}}+{a}{b}+{{b}}^{{2}}\right)}={a}\cdot{{a}}^{{2}}+{a}\cdot{a}{b}+{a}\cdot{{b}}^{{2}}-{b}\cdot{{a}}^{{2}}-{b}\cdot{a}{b}-{b}\cdot{{b}}^{{2}}=$

$={{a}}^{{3}}+{{a}}^{{2}}{b}+{a}{{b}}^{{2}}-{{a}}^{{2}}{b}-{a}{{b}}^{{2}}-{{b}}^{{3}}={{a}}^{{3}}-{{b}}^{{3}}$.

Similarly, it can be shown, that ${{a}}^{{3}}+{{b}}^{{3}}={\left({a}+{b}\right)}{\left({{a}}^{{2}}-{a}{b}+{{b}}^{{2}}\right)}$.

Or written more compactly: $a^3 \pm b^3=\left(a\pm b\right)\left(a^2 \mp ab+b^2 \right)$

Expressions ${{a}}^{{2}}+{a}{b}+{{b}}^{{2}}$ and ${{a}}^{{2}}-{a}{b}+{{b}}^{{2}}$ are often called incomplete squares, because they lack one ${a}{b}$ to become perfect square $\left({{\left({a}\pm{b}\right)}}^{{2}}={{a}}^{{2}}\pm{\color{red}{{{2}}}}{a}{b}+{{b}}^{{2}}\right)$.

Don't attempt to factor an incomplete square. It can't be factored.

Example 1. Factor ${{x}}^{{3}}+{8}$.

Notice, that ${8}={{2}}^{{3}}$.

Thus, ${{x}}^{{3}}+{8}={{x}}^{{3}}+{{2}}^{{3}}={\left({x}+{2}\right)}{\left({{x}}^{{2}}-{x}\cdot{2}+{{2}}^{{2}}\right)}={\left({x}+{2}\right)}{\left({{x}}^{{2}}-{2}{x}+{4}\right)}$.

Answer: ${{x}}^{{3}}+{8}={\left({x}+{2}\right)}{\left({{x}}^{{2}}-{2}{x}+{4}\right)}$.

Of course, there can be more complex expressions.

Example 2. Factor ${27}{{y}}^{{3}}-{64}$.

Notice, that ${27}{{y}}^{{3}}={{\left({3}{y}\right)}}^{{3}}$ and ${64}={{4}}^{{3}}$.

Thus, ${27}{{y}}^{{3}}-{64}={{\left({3}{y}\right)}}^{{3}}-{{4}}^{{3}}={\left({3}{y}-{4}\right)}{\left({{\left({3}{y}\right)}}^{{2}}+{3}{y}\cdot{4}+{{4}}^{{2}}\right)}={\left({3}{y}-{4}\right)}{\left({9}{{y}}^{{2}}+{12}{y}+{16}\right)}$.

Answer: ${27}{{y}}^{{3}}-{64}={\left({3}{y}-{4}\right)}{\left({9}{{y}}^{{2}}+{12}{y}+{16}\right)}$.

And even harder...

Example 3. Factor the following: ${8}{{m}}^{{6}}{{n}}^{{9}}+{27}{{a}}^{{9}}{{b}}^{{3}}$.

Notice, that ${8}{{m}}^{{6}}{{n}}^{{9}}={{\left({2}{{m}}^{{2}}{{n}}^{{3}}\right)}}^{{3}}$ and ${27}{{a}}^{{9}}{{b}}^{{3}}={{\left({3}{{a}}^{{3}}{b}\right)}}^{{3}}$.

Thus, ${8}{{m}}^{{6}}{{n}}^{{9}}+{27}{{a}}^{{9}}{{b}}^{{3}}={{\left({2}{{m}}^{{2}}{{n}}^{{3}}\right)}}^{{3}}+{{\left({3}{{a}}^{{3}}{b}\right)}}^{{3}}=$

$={\left({2}{{m}}^{{2}}{{n}}^{{3}}+{3}{{a}}^{{3}}{b}\right)}{\left({{\left({2}{{m}}^{{2}}{{n}}^{{3}}\right)}}^{{2}}-{2}{{m}}^{{2}}{{n}}^{{3}}\cdot{3}{{a}}^{{3}}{b}+{{\left({3}{{a}}^{{3}}{b}\right)}}^{{2}}\right)}=$

$={\left({2}{{m}}^{{2}}{{n}}^{{3}}+{3}{{a}}^{{3}}{b}\right)}{\left({4}{{m}}^{{4}}{{n}}^{{6}}-{6}{{a}}^{{3}}{b}{{m}}^{{2}}{{n}}^{{3}}+{9}{{a}}^{{6}}{{b}}^{{2}}\right)}$.

Answer: ${8}{{m}}^{{6}}{{n}}^{{9}}+{27}{{a}}^{{9}}{{b}}^{{3}}={\left({2}{{m}}^{{2}}{{n}}^{{3}}+{3}{{a}}^{{3}}{b}\right)}{\left({4}{{m}}^{{4}}{{n}}^{{6}}-{6}{{a}}^{{3}}{b}{{m}}^{{2}}{{n}}^{{3}}+{9}{{a}}^{{6}}{{b}}^{{2}}\right)}$.

Now, it is time to exercise.

Exercise 1. Factor the following: ${{n}}^{{3}}+{125}$.

Answer: ${\left({n}-{5}\right)}{\left({{n}}^{{2}}+{5}{n}+{25}\right)}$.

Exercise 2. Factor the following: $-{343}+{{y}}^{{6}}{{x}}^{{3}}$.

Answer: ${\left({{y}}^{{2}}{x}-{7}\right)}{\left({{y}}^{{4}}{{x}}^{{2}}+{7}{{y}}^{{2}}{x}+{49}\right)}$. Hint: $-{343}+{{y}}^{{6}}{{x}}^{{3}}={{y}}^{{6}}{{x}}^{{3}}-{343}$.

Exercise 3. Factor ${{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{27}{{a}}^{{3}}$.

Answer: ${\left({x}{y}{z}-{3}{a}\right)}{\left({{x}}^{{2}}{{y}}^{{2}}{{z}}^{{2}}+{3}{a}{x}{y}{z}+{9}{{a}}^{{2}}\right)}$.

Exercise 4. Factor ${8}{{x}}^{{3}}+{1}$.

Answer: ${\left({2}{x}+{1}\right)}{\left({4}{{x}}^{{2}}-{2}{x}+{1}\right)}$.