Hardest maths questions - solve equation involving cube roots - Free Printable

We are given the equation:

$$
\sqrt[3]{n + \sqrt{n^2 + 8}} + \sqrt[3]{n - \sqrt{n^2 + 8}} = 8
$$

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Let’s denote:

$$
a = \sqrt[3]{n + \sqrt{n^2 + 8}}, \quad b = \sqrt[3]{n - \sqrt{n^2 + 8}}
$$

So the equation becomes:

$$
a + b = 8 \tag{1}
$$

Now, let’s compute $ ab $. First, note that:

$$
ab = \sqrt[3]{(n + \sqrt{n^2 + 8})(n - \sqrt{n^2 + 8})}
$$

This is a difference of squares:

$$
(n + \sqrt{n^2 + 8})(n - \sqrt{n^2 + 8}) = n^2 - (n^2 + 8) = -8
$$

Therefore:

$$
ab = \sqrt[3]{-8} = -2 \tag{2}
$$

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Now we have:

- $ a + b = 8 $
- $ ab = -2 $

Recall the identity for cubes:

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

We can use this to find $ a^3 + b^3 $.

From definition:

$$
a^3 = n + \sqrt{n^2 + 8}, \quad b^3 = n - \sqrt{n^2 + 8}
$$

So:

$$
a^3 + b^3 = (n + \sqrt{n^2 + 8}) + (n - \sqrt{n^2 + 8}) = 2n \tag{3}
$$

Now plug into the identity:

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

Substitute known values:

Left side: $ 8^3 = 512 $

Right side: $ 2n + 3(-2)(8) = 2n - 48 $

So:

$$
512 = 2n - 48
$$

Solve for $ n $:

$$
2n = 512 + 48 = 560 \\
n = \frac{560}{2} = 280
$$

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✔ Final Answer: $ \boxed{280} $

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Verification:

Let’s verify by plugging $ n = 280 $ back into the original expression.

Compute:

$$
\sqrt{n^2 + 8} = \sqrt{280^2 + 8} = \sqrt{78400 + 8} = \sqrt{78408}
$$

Note: $ 280^2 = 78400 $, so $ \sqrt{78408} $ is slightly more than 280.

But we don’t need to compute the exact cube roots — we already used algebraic identities that are valid regardless of the specific value. Since our derivation was purely algebraic and consistent, and we arrived at a clean integer solution, it’s highly likely correct.

Alternatively, check numerically:

Let’s approximate:

$ \sqrt{78408} \approx 280.014285... $ (since $ (280 + x)^2 = 78400 + 560x + x^2 \approx 78408 \Rightarrow 560x \approx 8 \Rightarrow x \approx 8/560 = 1/70 \approx 0.014285 $)

So:

- $ a = \sqrt[3]{280 + 280.014285} \approx \sqrt[3]{560.014285} \approx \sqrt[3]{560} $
- $ b = \sqrt[3]{280 - 280.014285} \approx \sqrt[3]{-0.014285} \approx -0.242 $

Wait — this seems problematic. But actually, our algebraic steps are correct. The issue is that we must be careful: when we computed $ ab = -2 $, that’s exact. And $ a + b = 8 $, so even if one term is large and the other small, their sum is 8 and product is -2.

Actually, let’s compute $ a^3 + b^3 = 2n = 560 $, and from identity:

$$
(a + b)^3 = a^3 + b^3 + 3ab(a + b) = 560 + 3(-2)(8) = 560 - 48 = 512 = 8^3
$$

Perfect match. So algebraically, everything checks out.

The numerical approximation went wrong because I miscomputed $ b^3 = n - \sqrt{n^2 + 8} \approx 280 - 280.014 = -0.014 $, but then $ b = \sqrt[3]{-0.014} \approx -0.24 $, and $ a = \sqrt[3]{560.014} \approx 8.24 $, and indeed $ 8.24 - 0.24 = 8 $. So numerically it also checks out!

Thus, the solution is verified.

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✔ Final Answer: $ \boxed{280} $