Porisma

Worked solution

Apply fractional and negative index laws

Simplify the following: (a) 81/3⋅27−1/23−1/6⋅91/3\dfrac{8^{1/3}\cdot27^{-1/2}}{3^{-1/6}\cdot9^{1/3}}; (b) x1/3+x−2/3x4/3\dfrac{x^{1/3}+x^{-2/3}}{x^{4/3}}.

CJC · O Level revision · Example 7

(a) Use a common base for the powers of 3

Answer

29\dfrac29.

Rewrite 8, 27 and 9

(23)1/3(33)−1/23−1/6(32)1/3=2⋅3−3/23−1/6⋅32/3\begin{aligned}&\frac{(2^3)^{1/3}(3^3)^{-1/2}}{3^{-1/6}(3^2)^{1/3}}\\ & =\frac{2\cdot3^{-3/2}}{3^{-1/6}\cdot3^{2/3}}\end{aligned}

Use 8=2³, 27=3³ and 9=3², as in the source.

Subtract both denominator exponents

2⋅3−3/2−(−1/6)−2/3=2⋅3−2=29\begin{aligned}&2\cdot3^{-3/2-(-1/6)-2/3}\\ & =2\cdot3^{-2}\\ & =\frac29\end{aligned}

The negative denominator exponent changes sign when subtracted.

How did this part go?

(b) Divide each numerator term by the denominator

Answer

1x+1x2\dfrac1x+\dfrac1{x^2}, with x≠0x\ne0.

Split the fraction before combining exponents

x1/3x4/3+x−2/3x4/3=x1/3−4/3+x−2/3−4/3=x−1+x−2=1x+1x2\begin{aligned}&\frac{x^{1/3}}{x^{4/3}}+\frac{x^{-2/3}}{x^{4/3}}\\ & =x^{1/3-4/3}+x^{-2/3-4/3}\\ & =x^{-1}+x^{-2}\\ & =\frac1x+\frac1{x^2}\end{aligned}

The source subtracts 4/3 from each numerator exponent. On the real cube-root interpretation the original expression is defined for every real x≠0x\ne0; this domain qualification is supplied independently.

How did this part go?