Find the Term Independent of x in the Expansion of [(x+1)/(x^(2/3) − x^(1/3) + 1) − (x−1)/(x − x^(1/2))]¹⁰

A standard binomial-theorem problem combining algebraic simplification with fractional exponents, not tied to a specific exam paper.

Problem

Find the term independent of $x$ in the expansion of $\left[\dfrac{x+1}{x^{2/3}-x^{1/3}+1} – \dfrac{x-1}{x-x^{1/2}}\right]^{10}$.

Key insight. Both fractions look intimidating only because of the fractional exponents — but $x^{2/3}-x^{1/3}+1$ is exactly the “$a^2-ab+b^2$” factor from the sum-of-cubes identity with $a=x^{1/3}$, and $x-\sqrt x$ hides a similar difference-of-squares structure. Once both simplify, the whole bracket collapses to something small enough to binomial-expand directly.

Watch this explained step by step →  ·  More on the Shiwam’s Classes channel

Approach

Simplify the first fraction by recognising $x+1$ as a sum of cubes with $a=x^{1/3}$, $b=1$, so it factors against the given denominator. Simplify the second fraction by factoring $x-1$ as a difference of squares in terms of $\sqrt x$. Subtract the simplified results to get a clean binomial base, then apply the general term formula to find which term is independent of $x$.

Solution

Step 1 — Simplify the first fraction using sum of cubes

With $a=x^{1/3}$, $b=1$: $x+1 = a^3+b^3 = (a+b)(a^2-ab+b^2) = \left(x^{1/3}+1\right)\left(x^{2/3}-x^{1/3}+1\right)$

$$\frac{x+1}{x^{2/3}-x^{1/3}+1} = x^{1/3}+1$$

Step 2 — Simplify the second fraction using difference of squares

$$\frac{x-1}{x-\sqrt x} = \frac{(\sqrt x-1)(\sqrt x+1)}{\sqrt x(\sqrt x-1)} = \frac{\sqrt x+1}{\sqrt x} = 1+x^{-1/2}$$

Step 3 — Subtract the two simplified fractions

$$\left(x^{1/3}+1\right) – \left(1+x^{-1/2}\right) = x^{1/3}-x^{-1/2}$$

Step 4 — Write the general term of the binomial expansion

For $\left(x^{1/3}-x^{-1/2}\right)^{10}$, the general term is:

$$T_{r+1} = \binom{10}{r}\left(x^{1/3}\right)^{10-r}\left(-x^{-1/2}\right)^r = \binom{10}{r}(-1)^r\,x^{\frac{10-r}{3}-\frac{r}{2}}$$

Step 5 — Set the exponent to zero

$$\frac{10-r}{3}-\frac{r}{2} = 0$$

Multiplying through by 6:

$$2(10-r)-3r = 0 \implies 20-2r-3r=0 \implies 20=5r \implies r=4$$

Step 6 — Evaluate the term at r = 4

$$T_5 = \binom{10}{4}(-1)^4 = \binom{10}{4} = \frac{10\times9\times8\times7}{4\times3\times2\times1} = 210$$

Answer

Common mistakes

• Trying to binomial-expand the original expression directly, without simplifying the fractions first. The unsimplified expression isn’t in the form $(a+b)^{10}$ at all — the algebraic simplification in Steps 1–3 is what turns it into a genuine two-term binomial that the general term formula can be applied to.
• Losing the sign when combining $-\left(1+x^{-1/2}\right)$. Distributing the negative sign correctly gives $-1-x^{-1/2}$, and the constant $-1$ cancels with the $+1$ from Step 1’s $x^{1/3}+1$ — dropping this cancellation leaves an extra constant term that doesn’t belong in the binomial base.

Practise next

• Find the term independent of $x$ in the expansion of $\left[\dfrac{x+8}{x^{2/3}-2x^{1/3}+4} – \dfrac{x-4}{x-2\sqrt x}\right]^{9}$, using the same sum-of-cubes and difference-of-squares simplification approach.