Question

Prove that $${\left[{\displaystyle \frac{x^{a^2}}{x^{b^2}}}\right]}^{{\displaystyle \frac{1}{a+b}}}\times {\left[{\displaystyle \frac{x^{b^2}}{x^{c^2}}}\right]}^{{\displaystyle \frac{1}{b+c}}}\times {\left[{\displaystyle \frac{x^{c^2}}{x^{a^2}}}\right]}^{{\displaystyle \frac{1}{c+a}}}=1$$ .

Answer 1

Hint: To prove that that $${\left[{\displaystyle \frac{x^{a^2}}{x^{b^2}}}\right]}^{{\displaystyle \frac{1}{a+b}}}\times {\left[{\displaystyle \frac{x^{b^2}}{x^{c^2}}}\right]}^{{\displaystyle \frac{1}{b+c}}}\times {\left[{\displaystyle \frac{x^{c^2}}{x^{a^2}}}\right]}^{{\displaystyle \frac{1}{c+a}}}=1$$, we will consider the LHS. Using the formula ${\displaystyle \frac{a^m}{a^n}}=a^{m-n}$ , we can write the LHS as $${\left(x^{a^2-b^2}\right)}^{{\displaystyle \frac{1}{a+b}}}\times {\left(x^{b^2-c^2}\right)}^{{\displaystyle \frac{1}{b+c}}}\times {\left(x^{c^2-a^2}\right)}^{{\displaystyle \frac{1}{c+a}}}$$. Using the formula ${\left(a^m\right)}^n=a^{mn}$, we can further simplify the previous equation to get $$\left(x^{{{\displaystyle \frac{a^2-b}{a+b}}}^2}\right)\times \left(x^{{\displaystyle \frac{b^2-c^2}{b+c}}}\right)\times \left(x^{{\displaystyle \frac{c^2-a^2}{c+a}}}\right)$$ . Let us apply the identity $a^2-b^2=(a+b)(a-b)$ on the powers and simplify to get $$x^{(a-b)}\times x^{(b-c)}\times x^{(c-a)}$$ . Using the exponent rule $a^m\times a^n=a^{m+n}$ , we will get $$x^{a-b+b-c+c-a}$$ . On solving this and using $a^0=1$ , the LHS will be equal to RHS.

Complete step by step answer:
We have to prove that that $${\left[{\displaystyle \frac{x^{a^2}}{x^{b^2}}}\right]}^{{\displaystyle \frac{1}{a+b}}}\times {\left[{\displaystyle \frac{x^{b^2}}{x^{c^2}}}\right]}^{{\displaystyle \frac{1}{b+c}}}\times {\left[{\displaystyle \frac{x^{c^2}}{x^{a^2}}}\right]}^{{\displaystyle \frac{1}{c+a}}}=1$$ .
Let us consider the RHS.
We have $${\left[{\displaystyle \frac{x^{a^2}}{x^{b^2}}}\right]}^{{\displaystyle \frac{1}{a+b}}}\times {\left[{\displaystyle \frac{x^{b^2}}{x^{c^2}}}\right]}^{{\displaystyle \frac{1}{b+c}}}\times {\left[{\displaystyle \frac{x^{c^2}}{x^{a^2}}}\right]}^{{\displaystyle \frac{1}{c+a}}}$$
We know that ${\displaystyle \frac{a^m}{a^n}}=a^{m-n}$ . Hence, we can write the above equation as
$${\left(x^{a^2-b^2}\right)}^{{\displaystyle \frac{1}{a+b}}}\times {\left(x^{b^2-c^2}\right)}^{{\displaystyle \frac{1}{b+c}}}\times {\left(x^{c^2-a^2}\right)}^{{\displaystyle \frac{1}{c+a}}}$$
We know that ${\left(a^m\right)}^n=a^{mn}$ . Thus, the above equation can be written as
$$\left(x^{{{\displaystyle \frac{a^2-b}{a+b}}}^2}\right)\times \left(x^{{\displaystyle \frac{b^2-c^2}{b+c}}}\right)\times \left(x^{{\displaystyle \frac{c^2-a^2}{c+a}}}\right)$$
We know that $a^2-b^2=(a+b)(a-b)$ . We can express the powers in this form. We will get
$$\left(x^{{\displaystyle \frac{(a+b)(a-b)}{a+b}}}\right)\times \left(x^{{\displaystyle \frac{(b+c)(b-c)}{b+c}}}\right)\times \left(x^{{\displaystyle \frac{(c+a)(c-a)}{c+a}}}\right)$$
Let’s cancel the common terms from the numerator and denominator of the powers.
$$x^{(a-b)}\times x^{(b-c)}\times x^{(c-a)}$$
We know that $a^m\times a^n=a^{m+n}$ . Hence, the above equation becomes
$$x^{a-b+b-c+c-a}$$
When solving the powers, we will get
$$x^0$$
We know that $a^0=1$ . Thus,
$$x^0=1=RHS$$
Hence, $LHS=RHS$
That is, $${\left[{\displaystyle \frac{x^{a^2}}{x^{b^2}}}\right]}^{{\displaystyle \frac{1}{a+b}}}\times {\left[{\displaystyle \frac{x^{b^2}}{x^{c^2}}}\right]}^{{\displaystyle \frac{1}{b+c}}}\times {\left[{\displaystyle \frac{x^{c^2}}{x^{a^2}}}\right]}^{{\displaystyle \frac{1}{c+a}}}=1$$
Hence, proved.

Note: In these types of questions, we can see that the bases are the same. So we have to simplify the terms and exponents using the rules of exponents. Rules of exponents must be studied thoroughly to solve this question. You may make mistakes in the rule for ${\left(a^m\right)}^n$ by writing ${\left(a^m\right)}^n=a^{m+n}$ . Also, mistakes may be committed by writing ${\displaystyle \frac{a^m}{a^n}}=a^{m+n}$ and $a^m\times a^n=a^{m-n}$ . You may also write the identity $a^2-b^2$ as $=a^2-2ab+b^2$ which leads to improper results.