Question

maths(SM) ch1pg1.35 q9 Q.B.

PLZ explain in detail step by step

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Answer 1

$\mathsf{9.\;\;Given : \;\bigg[(x)^{\dfrac{b + c}{c - a}}\bigg]^{\dfrac{1}{a - b}} \times \bigg[(x)^{\dfrac{c + a}{a - b}}\bigg]^{\dfrac{1}{b - c}} \times \bigg[(x)^{\dfrac{a + b}{b - c}}\bigg]^{\dfrac{1}{c - a}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

Solution is attached above and is 1 ......