Math, asked by physicsloverhere, 1 month ago

\mathsf{9.\;\;  \;\bigg[(x)^{\dfrac{b + c}{c - a}}\bigg]^{\dfrac{1}{a - b}} \times \bigg[(x)^{\dfrac{c + a}{a - b}}\bigg]^{\dfrac{1}{b - c}} \times \bigg[(x)^{\dfrac{a + b}{b - c}}\bigg]^{\dfrac{1}{c - a}}} reduce to?


Answers

Answered by shikhakumari8743
7

Answer:

Similarly, a random matrix is a matrix whose elements are random variables. In particular, we can have an m by n random matrix M as

we

Answered by Anonymous
220

Given : -

  • \mathsf{\: \;\bigg[(x)^{\dfrac{b + c}{c - a}}\bigg]^{\dfrac{1}{a - b}} \times \bigg[(x)^{\dfrac{c + a}{a - b}}\bigg]^{\dfrac{1}{b - c}} \times \bigg[(x)^{\dfrac{a + b}{b - c}}\bigg]^{\dfrac{1}{c - a}}}

Need To Find Out :-

  • We have to find out the value of given question.

Solution :-

\mathsf\pink{{\: \;\bigg[(x)^{\dfrac{b + c}{c - a}}\bigg]^{\dfrac{1}{a - b}} \times \bigg[(x)^{\dfrac{c + a}{a - b}}\bigg]^{\dfrac{1}{b - c}} \times \bigg[(x)^{\dfrac{a + b}{b - c}}\bigg]^{\dfrac{1}{c - a}}}}\\\\

  • \;\;\textsf{We know that : \purple{\boxed{\mathsf{\big[(a)^m\big]^n = (a)^{mn}}}}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{b + c}{(c - a)(a - b)}\bigg]} \times (x)^{\bigg[\dfrac{c + a}{(a - b)(b - c)}\bigg]} \times  (x)^{\bigg[\dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

  • \;\;\textsf{We know that : \purple{\boxed{\mathsf{(a)^m \times(a)^n = (a)^{m + n}}}}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{b + c}{(c - a)(a - b)} + \dfrac{c + a}{(a - b)(b - c)} + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{b + c}{c - a} + \dfrac{c + a}{b - c}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{(b + c)(b - c) + (c + a)(c - a)}{(c - a)(b - c)}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{b^2 - c^2 + c^2 - a^2}{(c - a)(b - c)}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{b^2 - a^2}{(c - a)(b - c)}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{-(a^2 - b^2)}{(c - a)(b - c)}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{1}{a - b} \bigg(\dfrac{-(a - b)(a + b)}{(c - a)(b - c)}\bigg) + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{\bigg[\dfrac{-(a + b)}{(c - a)(b - c)} + \dfrac{a + b}{(b - c)(c - a)}\bigg]}}\\\\

\mathsf{:\implies (x)^{0}}\\\\

  • \;\;\textsf{We know that : \purple{\boxed{\mathsf{a^0 = 1}}}}\\\\

\mathsf\pink{{:\implies (x)^{0} = 1}}\\\\

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