# Does there exist a way to find the sum of the digits of the result from a large cubic root?

The problem is as follows:

Find the sum of the digits of $$B$$:

$$B=\sqrt[3]{1224 \times 1225\times 1226+35\sqrt[3]{34\times35\times36+35}}$$

The possible answers given in my book are as follows:

$$\begin{array}{ll} 1.&10\\ 2.&18\\ 3.&9\\ 4.&12\\ \end{array}$$

I've been tempted to use a calculator but this problem is meant to be solved by hand.

Does there exist a trick or anything for doing this, other than going by the brute force method of doing all the operations one by one?. How can this be simplified or solved quickly?. Can someone help me here?.

• Hint: $(a-1)a(a+1) + a = a^3$. Apply it twice. Mar 21, 2020 at 0:49
• As Catalin says, recall $(n-1)(n+1) = n^2 - 1.$ Multiply by an extra $n,$ we get $(n-1)n(n+1) = n^3 - n.$ Here $n$ is the middle of three numbers. Mar 21, 2020 at 0:57

Well, I suppose you could just do it by hand, but that would be a pain in the butt. Thus, to flesh out Catalin Zara's hint:

Let $$a=1225, b=35$$. Then

$$B = \sqrt[3]{(a-1)(a)(a+1) + b\sqrt[3]{(b-1)(b)(b+1) + b}}$$

Noting that $$(x-1)(x)(x+1) = (x^2 - 1)(x) = x^3 - x$$, then the innermost radicand is equal to $$b^3$$ -- just move the $$x$$ over in the previous equation. Then that simplifies to $$b$$ and thus

$$B = \sqrt[3]{(a-1)(a)(a+1) + b^2}$$

A priori, it doesn't seem like we can apply it again. However, would it not be convenient for us if $$b^2 = a$$, to reapply the same trick? Indeed, if you happen to have $$35^2$$ memorized, or calculate it by hand, you'll find $$b^2 = a$$. Then the radicand simplifies to $$a^3$$ and then $$B=a$$.

Thus, $$B=1225$$, and finding the sum of the digits is trivial.