Cube of Sum and Difference

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Cube of sum and difference:

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

Let's see how to derive it.

Recall, that exponent is just repeating multiplication.

Thus, we can write that $$${{\left({a}+{b}\right)}}^{{3}}={\left({a}+{b}\right)}{{\left({a}+{b}\right)}}^{{2}}$$$.

From square of sum/difference note, we know, that $$${{\left({a}+{b}\right)}}^{{2}}={{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}$$$.

Thus, $$${{\left({a}+{b}\right)}}^{{3}}={\left({a}+{b}\right)}{\left({{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}\right)}$$$.

Finally, just multiply polynomials: $$${\left({a}+{b}\right)}{\left({{a}}^{{2}}+{2}{a}{b}+{{b}}^{{2}}\right)}={a}\cdot{{a}}^{{2}}+{a}\cdot{2}{a}{b}+{a}\cdot{{b}}^{{2}}+{b}\cdot{{a}}^{{2}}+{b}\cdot{2}{a}{b}+{b}\cdot{{b}}^{{2}}=$$$

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

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

Or, more shortly: $$${{\left({a}\pm{b}\right)}}^{{3}}={{a}}^{{3}}\pm{3}{{a}}^{{2}}{b}+{3}{a}{{b}}^{{2}}\pm{{b}}^{{2}}$$$.

Example 1. Multiply $$${{\left({2}{x}+{y}\right)}}^{{3}}$$$.

Here $$${a}={2}{x}$$$ and $$${b}={y}$$$.

Just use above formula: $$${{\left({2}{x}+{y}\right)}}^{{3}}={{\left({2}{x}\right)}}^{{3}}+{3}\cdot{{\left({2}{x}\right)}}^{{2}}\cdot{\left({y}\right)}+{3}\cdot{\left({2}{x}\right)}\cdot{{\left({y}\right)}}^{{2}}+{{\left({y}\right)}}^{{3}}={8}{{x}}^{{3}}+{12}{{x}}^{{2}}{y}+{6}{x}{{y}}^{{2}}+{{y}}^{{3}}$$$.

Let's see how to handle minus sign.

Example 2. Multiply $$${{\left(\frac{{3}}{{4}}{a}{b}-{2}{c}{d}\right)}}^{{3}}$$$.

Here $$${a}=\frac{{3}}{{4}}{a}{b}$$$ and $$${b}={2}{c}{d}$$$.

Now, use formula for difference: $$${{\left(\frac{{3}}{{4}}{a}{b}-{2}{c}{d}\right)}}^{{3}}={{\left(\frac{{3}}{{4}}{a}{b}\right)}}^{{3}}-{3}\cdot{{\left(\frac{{3}}{{4}}{a}{b}\right)}}^{{2}}\cdot{\left({2}{c}{d}\right)}+{3}\cdot{\left(\frac{{3}}{{4}}{a}{b}\right)}\cdot{{\left({2}{c}{d}\right)}}^{{2}}-{{\left({2}{c}{d}\right)}}^{{3}}=$$$

$$$=\frac{{27}}{{64}}{{a}}^{{3}}{{b}}^{{3}}-\frac{{27}}{{8}}{{a}}^{{2}}{{b}}^{{2}}{c}{d}+{9}{a}{b}{{c}}^{{2}}{{d}}^{{2}}-{8}{{c}}^{{3}}{{d}}^{{3}}$$$.

Finally, let's do a slightly harder example.

Example 3. Multiply the following: $$${{\left(-{x}{y}{z}-{2}{{x}}^{{2}}\right)}}^{{3}}$$$.

Till now, we didn't see two minus signs, but this case can be handled easily.

There are two options:

  • $$${a}=-{x}{y}{z}$$$ and $$${b}=-{2}{{x}}^{{2}}$$$; apply sum formula.
  • $$${a}=-{x}{y}{z}$$$ and $$${b}={2}{{x}}^{{2}}$$$; apply difference formula.

I choose second option: $$${{\left(-{x}{y}{z}-{2}{{x}}^{{2}}\right)}}^{{3}}={{\left(-{x}{y}{z}\right)}}^{{3}}-{3}\cdot{{\left(-{x}{y}{z}\right)}}^{{2}}\cdot{\left({2}{{x}}^{{2}}\right)}+{3}\cdot{\left(-{x}{y}{z}\right)}\cdot{{\left({2}{{x}}^{{2}}\right)}}^{{2}}-{{\left({2}{{x}}^{{2}}\right)}}^{{3}}=$$$

$$$=-{{x}}^{{3}}{{y}}^{{3}}{{z}}^{{3}}-{6}{{x}}^{{4}}{{y}}^{{2}}{{z}}^{{2}}-{12}{{x}}^{{5}}{y}{z}-{8}{{x}}^{{6}}$$$.

From last example we see, that $$${\color{purple}{{{{\left(-{a}-{b}\right)}}^{{3}}=-{{\left({a}+{b}\right)}}^{{3}}}}}$$$.

Now, it is time to exercise.

Exercise 1. Multiply $$${{\left({4}{z}+{3}{y}\right)}}^{{3}}$$$.

Answer: $$${64}{{z}}^{{3}}+{144}{{z}}^{{2}}{y}+{108}{z}{{y}}^{{2}}+{27}{{y}}^{{3}}$$$.

Exercise 2. Multiply $$${{\left(-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}+{2}{x}\right)}}^{{3}}$$$.

Answer: $$$-\frac{{1}}{{27}}{{x}}^{{9}}{{y}}^{{6}}+\frac{{2}}{{3}}{{x}}^{{7}}{{y}}^{{4}}-{4}{{x}}^{{5}}{{y}}^{{2}}+{8}{{x}}^{{3}}$$$.

Hint: either swap summands ($$${{\left(-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}+{2}{x}\right)}}^{{3}}={{\left({2}{x}-\frac{{1}}{{3}}{{x}}^{{3}}{{y}}^{{2}}\right)}}^{{3}}$$$: commutative property of addition) or proceed as always.

Exercise 3. Multiply the following: $$${{\left(-{2}{x}-{1}\right)}}^{{3}}$$$.

Answer: $$$-{8}{{x}}^{{3}}-{12}{{x}}^{{2}}-{6}{x}-{1}$$$.