I have used change of variable while trying to solve quadratic equations. Recently, I came across some very clever uses of this technique. Some of the examples below served me as a reminder that it is OK to arbitrarily change the value of one side of an equation… as long as we compensate for the change!
Example #1
\[\text{Find } m^{3} + \frac{1}{m^{3}} \text{ if } m + \frac{1}{m} = 2\]
Solution: \(\boldsymbol{m = 1}\). Therefore \(\boldsymbol{m^{3} + \frac{1}{m^{3}} = 1^{3} + \frac{1}{1^{3}} = 1 + 1 = 2}\).
Method #1 – The obvious way is to solve for \(\boldsymbol m \):
\begin{align}
m + \frac{1}{m} &= 2\\
m^{2} + 1 &= 2m\\
m^{2} – 2m + 1 &= 0\\
\left( m – 1 \right)^{2} &= 0\\
m &= 1
\end{align}
Method #2 – Not necessary on this problem, but worthwhile looking into, because it will serve as a template for solving other problems.
Background: Every algebra student should know by heart the Sum of Cubes formula: \(x^{3} + 1 = \left(x + 1\right) \left(x^{2} – x + 1\right)\). Likewise, \(x^{3} + y^{3} = \left(x + y\right) \left(x^{2} – xy + y^{2}\right)\).
Knowing this, we substitute: \(\large \boldsymbol{x = m}, \text{ and } \boldsymbol{y = \frac{1}{m}}\).
\[\begin{align}
\text{This: } x^{3} + y^{3} &= \left(x + y\right) \left(x^{2} – xy + y^{2}\right) \\
\text{Becomes this: } m^{3} + \frac{1}{m^{3}} &= \left(m + \frac{1}{m}\right)\left(m^{2} – m\cdot\frac{1}{m} + \frac{1}{m^{2}}\right) \\
&= \left(m + \frac{1}{m}\right)\left(m^{2} + \frac{1}{m^{2}}\, – 1\right) \\
\text{We know the value of } \left(m + \frac{1}{m}\right), \text{therefore: } &= 2\cdot\left(m^{2} + \frac{1}{m^{2}}\, – 1\right)
\end{align}\]
The task now is to convert \(\boldsymbol{2\cdot\left(m^{2} + \frac{1}{m^{2}} – 1 \right)}\) into some form of \(\boldsymbol{2\cdot\left(m + \frac{1}{m}\right)^{?}}\).
Because what we’re converting from has square values, the obvious power to use is square. Let’s try \(\boldsymbol{\left(m + \frac{1}{m}\right)^{2}}\). This expands to: \(\boldsymbol{\left(m^{2} + \frac{1}{m^{2}} + 2\right)}\). The two constants differ in value by a total of \(\boldsymbol{3}\). That is not a problem… as long as we compensate for the difference!
\[\begin{align}
\text{Continuing: } 2\cdot \left(m^{2} + \frac{1}{m^{2}}\, – 1\right) &= 2\cdot \left(\left(m + \frac{1}{m}\right)^{2} – 3 \right) \text{ compensating for the value of the constant,} \\
&= 2\cdot\left(2^{2} – 3\right) \\
&= 2\cdot 1 \\
\text{We get the same result: } m^{3} + \frac{1}{m^{3}} &= 2
\end{align}\]
Isn’t math just beautiful? 😎 After having this technique under our belt, we can re-use it to work out the second example.
Example #2
\[\text{Find } m^{4} + \frac{1}{m^{4}} \text{ if } m + \frac{1}{m} = 4\]
Background: Every algebra student knows the formula for Difference of Squares, very useful in factoring; for example (to make it more clear, I will use a letter instead of a constant): \(\boldsymbol{x^{2} – y^{2} = \left(x + y\right)\left(x\, – y\right)}\). What may not be used as often is the Sum of Squares: \(\boldsymbol{x^{2} + y^{2} = \left(x + y\right)^{2}\, – 2x y}\).
Proceeding in the same manner as Method #2 above, substitute: \(\large \boldsymbol{m^{2}} \text{ for } \boldsymbol{x} \text{ and } \boldsymbol{\frac{1}{m^{2}}} \text{ for } \boldsymbol{y}\).
\[ \begin{align}
\left(m^{2}\right)^{2} + \left(\frac{1}{m^{2}}\right)^{2} &= \left(m^{2} + \frac{1}{m^{2}}\right)^{2} \, – 2\,m^{2}\cdot\frac{1}{m^{2}} \\
&= \left(m^{2} + \frac{1}{m^{2}}\right)^{2} \, – 2
\end{align} \]
As before, the task now is no convert \(\boldsymbol{\left(m^{2} + \frac{1}{m^{2}}\right)^{2}\, – 2}\) into some form of \(\boldsymbol{\left(m + \frac{1}{m}\right)}\). Enclosing the statement within a square overshoots by the constant \(\boldsymbol{2}\), which means we can compensate and subtract it at the end.
\[ \begin{align}
\text{Continuing: } \left(m^{2} + \frac{1}{m^{2}}\right)^{2} – 2 &= \left(\left(m + \frac{1}{m}\right)^{2} \, – 2 \right)^{2} – 2 \\
&= \left(4^{2} – 2\right)^{2} – 2 \\
&= 14^{2} – 2 \\
&= 196 – 2 \\
&= 194
\end{align} \]
I think that change of variable, coupled with well-known pre-defined formulas and the confidence to modify one side of an equation provided the change is compensated, this combination is very powerful.
If you find any errors in this post, or have some thoughts about it, please, email me at [email protected] .