Question

full solution send kro
show that kro​

Answer 1

Step-by-step explanation:

$\sf\large\underline\pink{Answer:-}$

$\displaystyle \sf \star \: {(\frac{3a}{5} - \frac{5b}{3} )}^{2} + 2ab = \frac{ {9a}^{2} }{25} + \frac{ {25b}^{2} }{9}$

From LHS

$\displaystyle \sf \star \: {(\frac{3a}{5} - \frac{5b}{3} )}^{2} + 2ab = \frac{ {9a}^{2} }{25} + \frac{ {25b}^{2} }{9} \\ \\ \displaystyle \sf \implies\frac{ {9a}^{2} }{25} + \frac{ {25b}^{2} }{9} - 2 \frac{3a}{5} \frac{5b}{3} + 2ab \: \: \\ \\ \displaystyle \sf \implies \: \frac{ {9a}^{2} }{25} + \frac{ {25b}^{2} }{9} - \frac{30ab}{15} + 2ab \: \: \: \\ \\ \displaystyle \sf \implies \: \frac{ {9a}^{2} }{25} + \frac{ {25b}^{2} }{9} - \frac{30ab + 30ab}{15} \\ \\ \displaystyle \sf \implies \: \boxed{ \underline{ \bold{ \frac{ {9a}^{2} }{25} + \frac{ {2b}^{2} }{9} }}}$

$\sf\large\underline\pink{Answer\:\:\:2:-}$

$\displaystyle \sf \star \: ( {x}^{2} - {y}^{2} )( {x}^{2} + {y}^{2} ) + ( {y}^{2} - {z}^{2} ) ( {y}^{2} + {z}^{2} ) + ( {z}^{2} - {x}^{2} ) ( {z}^{2} + {x}^{2} ) = 0$

From LHS

$\displaystyle \sf \star \: ( {x}^{2} - {y}^{2} )( {x}^{2} + {y}^{2} ) + ( {y}^{2} - {z}^{2} ) ( {y}^{2} + {z}^{2} ) + ( {z}^{2} - {x}^{2} ) ( {z}^{2} + {x}^{2} ) = 0 \\ \\ \displaystyle \sf \implies \: ( {x}^{4} - {y}^{4} ) + ( {y}^{4} - {z}^{4} ) + ( {z}^{4} - {x}^{4} ) \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \\ \displaystyle \sf \implies {x}^{4} - {y}^{4} + {y}^{4} - {z}^{4} + {z}^{4} - {x}^{4} = 0 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$