Find the cube root of the following rational numbers: -19683 2468… - Vidyadip

Question

Find the cube root of the following rational numbers: $\frac{-19683}{24689}$

✓

Answer

Let us consider the following rational number: $\frac{10648}{12167}$ Now $\sqrt[3]{\frac{-19689}{24389}}$
$=\frac{\sqrt[3]{-19689}}{\sqrt[3]{24689}}$
$\Big(\because\sqrt[3]{\frac{\text{a}}{\text{}b}}=\sqrt[3]{\frac{\sqrt[3]{\text{a}}}{\sqrt[3]{\text{b}}}}\Big)$
$=\frac{-\sqrt[3]{19689}}{\sqrt[3]{24689}}$
$\Big(\because\sqrt[3]{\text{-a}}=-\sqrt[3]{\text{a}} \Big)$
Cube root by factors: On factorising $19683$ into prime factors,
we get: $19683$ $= 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3 \times 3$
On grouping the factors in triples of equal factors,
we get: $19683 = {3 \times 3 \times 3} \times {3 \times 3 \times 3} \times {3 \times 3 \times 3}$
Now, taking one factor from each triple,
we get: $\sqrt[3]{19683 }$ $= 3 \times 3 \times 3 = 27$
Also On grouping the factors in triples of equal factors,
we get: $24389 = 29 \times 29 \times 29$ Now, taking one factor from the triple,
we get: $24389 = {29 \times 29 \times 29}$ Now, taking one factor from each triple,
we get: $\sqrt[3]{24389}=29$ Now $\sqrt[3]\frac{-19683}{24389}$
$=\frac{\sqrt[3]{-19689}}{\sqrt[3]{24689}}$
$=\frac{-\sqrt[3]{19689}}{\sqrt[3]{24689}}$
$=\frac{-27}{29}$

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