Question
Simplify the following:$\left(\frac{64 a^{12}}{27 b^6}\right)^{-\frac{2}{3}}$
✓
Answer
$\left(\frac{64 a^{12}}{27 b^6}\right)^{-\frac{2}{3}}$
$=\left(\frac{2^6 a^{12}}{3^3 b^6}\right)^{-\frac{2}{3}}$
$=\left(\frac{2^{6 \times\left(-\frac{2}{3}\right)} a^{12 \times\left(-\frac{2}{3}\right)}}{3^{3 \times\left(-\frac{2}{3}\right)} b^{6 \times\left(-\frac{2}{3}\right)}}\right) \ldots \ldots($Using $\left(a \times b^n\right)=a^n \times b^n$ and $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n})$
$=\frac{2^{-4} a^{-8}}{3^{-2} b^{-4}}$
$=\frac{3^2 b^4}{2^4 a^8} \cdots \cdot($Using $a^{-n}=\frac{1}{a^n})$
$=\frac{9 b^4}{16 a^8} .$
$=\left(\frac{2^6 a^{12}}{3^3 b^6}\right)^{-\frac{2}{3}}$
$=\left(\frac{2^{6 \times\left(-\frac{2}{3}\right)} a^{12 \times\left(-\frac{2}{3}\right)}}{3^{3 \times\left(-\frac{2}{3}\right)} b^{6 \times\left(-\frac{2}{3}\right)}}\right) \ldots \ldots($Using $\left(a \times b^n\right)=a^n \times b^n$ and $\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n})$
$=\frac{2^{-4} a^{-8}}{3^{-2} b^{-4}}$
$=\frac{3^2 b^4}{2^4 a^8} \cdots \cdot($Using $a^{-n}=\frac{1}{a^n})$
$=\frac{9 b^4}{16 a^8} .$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The cost of production of a video game is $Rs.5200.$ This is divided between material, labour and overheads in the ratio $5:6:2.$ If the video game is marked at a price that gives a $30\%$ profit, find the marked price. If the cost of material, labour and overheads increased by $40\%, 30\%$ and $10\%$ respectively, calculate the cost of manufacturing the video game now and the marked price so as to get the same percentage as before.
→Find the amount and compound interest on $Rs.7500$ for $1 \frac{1}{2}$ years at $8 \%$, payable semi$-$annually.
→There are two examination halls A and B. If 12 pupils are sent from A to B, the number of pupils in each room becomes the same. If 11 pupils are sent from room B to room A, then the number of pupils in A is double their number in B. Find the number of pupils in each room.
Given below are the marks obtained by $30$ students in an examination:
Taking class intervals $1 - 10, 11 - 20,,\dots ....., 41 - 50;$make a frequency table for the above distribution.
→| $08$ | $17$ | $33$ | $41$ | $47$ | $23$ | $20$ | $34$ |
| $09$ | $18$ | $42$ | $14$ | $30$ | $19$ | $29$ | $11$ |
| $36$ | $48$ | $40$ | $24$ | $22$ | $02$ | $16$ | $21$ |
| $15$ | $32$ | $47$ | $44$ | $33$ | $01$ |
In the given figure, a square is inscribed in a circle with center $O$. Find:
→- $∠BOC$
- $∠OCB$
- $∠COD$
- $∠BOD$
Is $BD$ a diameter of the circle?
Simplify the following :$\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$
→Solve the following pairs of equations$:\ \frac{x}{3}+\frac{y}{4}=11,\frac{5 x}{6}-\frac{y}{3}=-7$
→Find the area of a rhombus whose perimeter is $260\ cm$ and the length of one of its diagonal is $66\ cm.$
→Equal sides $AB$ and $AC$ of an isosceles $\triangle ABC$ are produced. The bisectors of the exterior angle so formed meet at $D$. Prove that $AD$ bisects $\angle A.$
→→