Question

Prove It -

$\large \pink{\frac{ \rm(x^{a} ) ^{ \frac{1}{bc} } {(x}^{b} ) ^{ \frac{1}{ca} } (x ^{c} ) ^{ \frac{1}{ab} }} { \rm \sqrt[abc]{(x ^{a} ) ^{a} (x ^{b} )^{b}(x^{c})^{c} } }} \blue= \blue1$

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Answer 1

Solution:-

To prove:-

$\sf{\dfrac{\bigg(x^{a}\bigg)^{\dfrac{1}{bc}}\bigg(x^{b}\bigg)^{\dfrac{1}{ca}}\bigg(x^{c}\bigg)^{\dfrac{1}{ab}}}{\sqrt[abc]{\bigg(x^{a}\bigg)^{a}\bigg(x^{b}\bigg)^{b}\bigg(x^{c}\bigg)^{c}}}\;=\;1}$

Taking LHS,

$\sf{= \dfrac{\bigg(x^{a}\bigg)^{\dfrac{1}{bc}}\bigg(x^{b}\bigg)^{\dfrac{1}{ca}}\bigg(x^{c}\bigg)^{\dfrac{1}{ab}}}{\sqrt[abc]{\bigg(x^{a}\bigg)^{a}\bigg(x^{b}\bigg)^{b}\bigg(x^{c}\bigg)^{c}}}}$

Simplify the expression using (aᵐ)ⁿ = aᵐⁿ.

$\sf{= \dfrac{x^{\frac{a}{bc}}\times x^{\frac{b}{ca}}\times x^{\frac{c}{ab}}}{\sqrt[abc]{\bigg(x^{a}\bigg)^{a}\bigg(x^{b}\bigg)^{b}\bigg(x^{c}\bigg)^{c}}}}$

Calculate the product using aᵐ × aⁿ = aᵐ⁺ⁿ.

$\sf{= \dfrac{x^{\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}}}{\sqrt[abc]{\bigg(x^{a}\bigg)^{a}\bigg(x^{b}\bigg)^{b}\bigg(x^{c}\bigg)^{c}}}}$

$\sf{= \dfrac{x^{\frac{a^{2}+b^{2}+c^{2}}{abc}}}{\sqrt[abc]{\bigg(x^{a}\bigg)^{a}\bigg(x^{b}\bigg)^{b}\bigg(x^{c}\bigg)^{c}}}}$

Simplify the expression using (aᵐ)ⁿ = aᵐⁿ.

$\sf{= \dfrac{x^{\frac{a^{2}+b^{2}+c^{2}}{abc}}}{\sqrt[abc]{x^{a^{2}}\times x^{b^{2}}\times x^{c^{2}}}}}$

Calculate the product using aᵐ × aⁿ = aᵐ⁺ⁿ.

$\sf{= \dfrac{x^{\frac{a^{2}+b^{2}+c^{2}}{abc}}}{\sqrt[abc]{x^{a^{2}+b^{2}+c^{2}}}}}$

Use $\sf \sqrt[n]{a^{m}}=a^{\frac{m}{n}}$ to transform the expression.

$\sf{= \dfrac{x^{\frac{a^{2}+b^{2}+c^{2}}{abc}}}{x^{\frac{a^{2}+b^{2}+c^{2}}{abc}}}}$

We know that any expression divided by itself equals 1.

$\sf = 1$

LHS = RHS

Hence, proved.