If x = 10 + 6√3, Find the Value of x^(1/3) − 1

A standard surds simplification problem, not tied to a specific exam paper.

Problem

If $x=10+6\sqrt3$, find the value of $x^{1/3}-1$.

Key insight. A surd like $10+6\sqrt3$ that’s meant to have a clean cube root almost always comes from cubing a simple expression of the form $a+\sqrt b$ — assuming that form and expanding $(a+\sqrt b)^3$ turns the problem into matching a rational part and an irrational part separately.

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Approach

Assume $x^{1/3}=a+\sqrt b$ for some numbers $a$ and $b$ to be found. Cube both sides to get $x=(a+\sqrt b)^3$, expand it fully, and separate the result into a rational part and a part multiplying $\sqrt b$. Matching these against the rational part (10) and the irrational part ($6\sqrt3$) of $x$ gives two equations that pin down $a$ and $b$.

Solution

Step 1 — Assume the cube root has the form a + √b

$$x^{1/3} = a+\sqrt b \implies x = (a+\sqrt b)^3$$

Step 2 — Expand (a + √b)³

$$(a+\sqrt b)^3 = a^3+3a^2\sqrt b+3a b+b\sqrt b = \left(a^3+3ab\right) + \left(3a^2+b\right)\sqrt b$$

Step 3 — Match with x = 10 + 6√3

Comparing this to $10+6\sqrt3$ suggests $\sqrt b=\sqrt3$, so $b=3$. Matching the two parts:

$$a^3+3ab = 10 \quad \text{(rational part)}$$

$$3a^2+b = 6 \quad \text{(coefficient of }\sqrt3\text{)}$$

Step 4 — Solve for a using the second equation

With $b=3$:

$$3a^2+3=6 \implies 3a^2=3 \implies a^2=1 \implies a=1$$

(taking the positive root, since $x^{1/3}$ should be positive)

Step 5 — Verify with the first equation

$$a^3+3ab = 1^3+3(1)(3) = 1+9 = 10 \checkmark$$

Both equations are satisfied, confirming $a=1$, $b=3$.

Step 6 — Compute x^(1/3) − 1

$$x^{1/3} = a+\sqrt b = 1+\sqrt3$$

$$x^{1/3}-1 = (1+\sqrt3)-1 = \sqrt3$$

Answer

Common mistakes

• Guessing $b$ incorrectly. It’s tempting to try $b=12$ or another value that also produces a $\sqrt3$-like term (since $\sqrt{12}=2\sqrt3$) — but the cleanest match, and the one consistent with both equations simultaneously, is $b=3$.
• Forgetting to check both equations. Solving only $3a^2+b=6$ gives $a=1$ when $b=3$, but this value must also satisfy $a^3+3ab=10$ — skipping this verification risks accepting a value of $a$ that only happens to satisfy one of the two conditions.
• Taking the negative root for a. $a^2=1$ has two solutions, $a=1$ and $a=-1$, but only $a=1$ gives a positive cube root consistent with $x=10+6\sqrt3$ being a positive number greater than 1.

Practise next

• If $x=16+8\sqrt5$, find the value of $x^{1/3}-1$.