Question

(x^a+b)² × (x^b+c)² × (x^c+a)² / (x^a×x^b×x^c)⁴= 1​

Answer 1

Step-by-step explanation:

(xa+b)2(xb+c)2(xc+a)2÷( xaxbxc)4=1. 2 ... Simplify [p^(a-b)]^c x [p^(a-c)] ^b x [p^(-a)]^a x [p^b]^c ... x/6+y/15=4 how?

Answer 2

Step-by-step explanation:

$\bf \underline{Solution-}$

$\textsf{We have,}$

$\bf{LHS} \: \sf{ = \frac{( {x}^{a + b } {)}^{2} \times ( {x}^{b + c} {)}^{2} \times ( {x}^{c + a} {)}^{2} }{( {x}^{a} \times {x}^{b} \times {x}^{c} {)}^{4} } }$

$\sf{ = \frac{ {x}^{2(a + b)} \times {x}^{(2b + c)} \times {x}^{2(c + a)} }{ {x}^{4(a + b + c)} } }$

$\sf{ = \frac{ {x}^{2a +2 b} \times {x}^{2b +2 c} \times {x}^{2c + 2a} }{ {x}^{4a +4 b + 4c} } }$

$\sf{ = \frac{ {x}^{2a + 2b + 2b + 2c + 2c + 2a} }{ {x}^{4a + 4b + 4c} } }$

$\sf{ = \frac{ {x}^{4a + 4b + 4c} }{ {x}^{4a + 4b + 4c} } }$

$\sf{ = 1} \: \bf{RHS}$

$\bf \underline{Hence \:proved.}$