Maths / Cube and Cube Roots / Cube Root by Prime Factorisation

QUESTION

Evaluate the following:

 1 $\dpi{120} \fn_jvn \large 3\sqrt{\frac{0.027}{0.008}}\div 3\sqrt{\frac{0.729}{0.512}}-\frac{1}{3}$ 2 $\dpi{120} \fn_jvn \large 3\sqrt{3343}+3\sqrt{0.064}-3\sqrt{0.125}$ 3 $\dpi{120} \fn_jvn \large \left [\left ( 3\sqrt{\frac{-216}{42875}}+3\sqrt{\frac{64}{125}}\right ) \right ]\times 3\sqrt{\frac{343}{1331}}$
 OPTIONS A. (i) - 1, (ii) - 6.9 , (iii) - 2/5 B. (i) - 3 , (ii) - 7.1 , (iii) - 1/5 C. (i) - 4 , (ii) - 7.9 , (iii) - 2/5 D. (i) - 1 , (ii) -  6.5 , (iii) - 1/5
Right Option : A

EXPLANATION
Explain TypeExplanation Content
Text

We have :

$(i)3\sqrt{\frac{0.027}{0.008}}\div 3\sqrt{\frac{0.729}{0.512}}-\frac{1}{3}$

= $3\sqrt{\frac{0.3\times 0.3\times 0.3}{0.2\times 0.2\times 0.2}}\div 3\sqrt{\frac{0.9\times 0.9 \times 0.9}{0.8 \times 0.8 \times 0.8}}-\frac{1}{3}$

= $\left ( \frac{0.3}{0.2} \div \frac{0.9}{0.8}\right )-\frac{1}{3}=\left ( \frac{0.3}{0.2}\times \frac{0.8}{0.9} \right )-\frac{1}{3}=\frac{4}{3}-\frac{1}{3}=1$

We have,

$(ii)3\sqrt{3343}+3\sqrt{0.064}-3\sqrt{0.125}$

= $3\sqrt{7\times 7 \times 7}+3\sqrt{0.4\times 0.4\times 0.4}-3\sqrt{0.5\times 0.5\times 0.5}$

= $7+0.4-0.5=7-0.1=6.9$

We have

$(iii)\left [\left ( 3\sqrt{\frac{-216}{42875}}+3\sqrt{\frac{64}{125}}\right ) \right ]\times 3\sqrt{\frac{343}{1331}}$

= $\left (\frac{(-6)\times (-6)\times (-6)}{35\times 35\times 35} \right ) ^{\frac{1}{3}}+\left ( \frac{4\times 4\times 4}{5\times 5\times 5} \right )^{\frac{1}{3}}\times \left ( \frac{7\times 7\times 7}{11\times 11\times 11} \right )^{\frac{1}{3}}$

=$\left ( \frac{-6}{35}+\frac{4}{5} \right )\times \left ( \frac{7}{11} \right )=\left ( \frac{-6\div28}{35} \right )\times \frac{7}{11}=\frac{22}{35}\times \frac{7}{11}=\frac{2}{5}$

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