Question

answer 1st and 2nd question fast...
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Answer 1

$\sf \Large Solution!!$

$\sf \to \dfrac{x^{a+b}\times x^{b+c}\times x^{c+a}}{(x^{a}\times x^{b}\times x^{c})^{2}}=1$

$\sf \to \dfrac{x^{a+b+b+c+c+a}}{(x^{a}\times x^{b}\times x^{c})\times (x^{a}\times x^{b}\times x^{c})}=1$

$\sf \to \dfrac{x^{2a+2b+2c}}{x^{a+b+c+a+b+c}}=1$

$\sf \to \dfrac{\cancel{x^{2a+2b+2c}}}{\cancel{x^{2a+2b+2c}}}=1$

$\sf \to 1=1$

LHS = RHS

Hence, proved.

———————————————————

$\sf \to \dfrac{5(1-x)+3(1+x)}{1-2x}=8$

$\sf \to 5(1-x)+3(1+x)=8(1-2x)$

$\sf \to 5-5x+3+3x=8-16x$

$\sf \to 8-2x=8-16x$

$\sf \to 16x-2x=8-8$

$\sf \to 14x=0$

$\sf \to x=0$

Answer 2

Answer:

$\huge→(xa×xb×xc)2xa+b×xb+c×xc+a=1</p><p></p><p> \huge\sf \to \dfrac{x^{a+b+b+c+c+a}}{(x^{a}\times x^{b}\times x^{c})\times (x^{a}\times x^{b}\times x^{c})}=1→(xa×xb×xc)×(xa×xb×xc)xa+b+b+c+c+a=1</p><p></p><p> \huge\sf \to \dfrac{x^{2a+2b+2c}}{x^{a+b+c+a+b+c}}=1→xa+b+c+a+b+cx2a+2b+2c=1</p><p></p><p>=1</p><p></p><p> \huge\sf \to \dfrac{\cancel{x^{2a+2b+2c}}}{\cancel{x^{2a+2b+2c}}}=1→x2a+2b+2cx2a+2b+2c=1</p><p></p><p>\sf \to 1=1→1=1</p><p></p><p>LHS = RHS</p><p></p><p>Hence, proved.</p><p></p><p>$

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