Question

15. Prove that :
a + b + c / (a^-1*b^-1 )+( b^-1*c^-1) + (c^-1*a^- 1 ) = abc


pls tell it I will it as excellent​

Answer 1

Answer:

Prove:

a+b+c / (a^-1b^-1+b^-1c^-1+c^-1a^-1)=abc

Proof:

Taking LHS

Following Steps:

a+b+c/(1/ab+1/bc+1/ca)

a+b+c/(c+a+b)/abc

On taking reciprocal

a+b+c ×.abc/c+a+b

abc=RHS

Hence proved

Answer 2

$\large\boxed{\underline{\mathcal{\red{Q}\green{u}\pink{e}\orange{s}\blue{ti}\red{on.??}}}}$

Prove that :----

$\frac{a + b + c}{( {a}^{ - 1} {b}^{ - 1}) + ( {b}^{ - 1} {c}^{ - 1} + ( {c}^{ - 1} {a}^{ - 1} )}$

$\Large\bold\star\underline{\underline\textbf{Formula\:used}}$

• a^(-1) = 1/a
• b^(-1) = 1/b
• c^(-1) = 1/c

$\large\boxed{\underline{\mathcal{\red{A}\green{n}\pink{s}\orange{w}\blue{e}\red{r}}}}$

using this in denominator we get,,,,

$\red\leadsto \: \green{\frac{a + b + c}{ \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca} } } \\ \\ \red\leadsto \: \pink{\frac{a + b + c}{( \frac{a + b + c}{abc} )}} \\ \\ \red\leadsto \blue{ \cancel{(a + b + c)} \times \frac{abc}{ \cancel{(a + b + c)}} } \\ \\ \red\leadsto \: \red{\large\boxed{\bold{abc}}}$

$\textbf{Hence, Proved}$

$\large\underline\textbf{Hope it Helps You.}$