2260.31 – Outrageous Cube Root

Find the cube root of $4^{15} + 6^9 \cdot 2^{11} + 6^{10} \cdot 3^7 + 3^{24}$ .

Solution

This won't fit in your calculator. You need to know the formula for a binomial cube:

$(a+b)^3= a^3 + 3a^2b + 3ab^2 + b^3.$

One clue that this is needed is that so many of the exponents in the given quantity are multiples of 3. The next thing to do is to express all the factors in terms of just two quantities (the $a$ and $b$ of the formula). These will be the primes 2 and 3. Thus,

\begin{aligned} 4^{15} + 6^9 \cdot 2^{11} + 6^{10} \cdot 3^7 + 3^{24} &=& 2^{30} + 2^{20} 3^{9} + 2^{10} 3^{17} + 3^{24} \\ &=& 2^{30} + 3 \cdot 2^{20} 3^{8} + 3 \cdot 2^{10} 3^{16} + 3^{24} \\ &=& (2^{10})^3 + 3 \cdot (2^{10})^2 3^{8} + 3 \cdot 2^{10} (3^{8})^2 + (3^{8})^3 \\ &=& (2^{10} + 3^8)^3. \end{aligned}

Therefore, the cube root is

$\sqrt[3]{(2^{10} + 3^8)^3} = 2^{10} + 3^8 = 1024 + 6561 = 7585.$

Now you know something nice about the number 7585.