Question

$\sf If \: a^4 \: + \: b^4 \: + \: c^4 \: + \: d^4 \: = \: 4abcd$

$\sf Prove \: that \: !!$

$\sf \implies a \: = \: b \: = \: c \: = \: d$

Answer 1

Hey, mate,!!

Here's your answer,....

I am sorry, but I don't know the procedure method,....

But I know the substitution,....

U can substitute any value of a,b,c,d satisfying the condition,.. a=b=c=d

For example, I take abcd values as 2, it becomes

16+16+16+16=4×2×2×2×2
64 =4×16
64=64

Hope it helped you,....
if u wish u can mark me as the brainliest,...only if it has helped you,....

Answer 2

$\underline{\underline{\mathfrak{\Large{Solution : }}}}$



$\underline{\textsf{Given,}} ,\\ \\ \sf \implies a^4 \: + \: b^4 \: + \: c^4 \: + \: d^4 \: = \: 4abcd$





$\underline{\mathsf{Subtract \: 2a^2b^2 \: and \: 2c^2d^2 \: from \: both \: sides , }} \\ \\ \sf \implies a^4 \: + \: b^4 \: + \: c^4 \: + \: d^4 \: - \: 2a^2b^2 \: - \: 2c^2d^2 \: = \: 4abcd \: - \: 2 {a}^{2} {b}^{2} \: - \: 2 {c}^{2} {d}^{2}$





$\underline{\textsf{Rearranging the terms : }} \\ \\ \sf \implies a^4 \: + \: b^4 \: - \: 2a^2b^2 \: + \: c^4 \: + \: d^4 \: - \: 2c^2d^2 \: = \: -2( a^2b^2 \: + \: c^2d^2 \: - \: 2abcd ) \\ \\ \\ \sf \implies \{( {a}^{2} )^{2} \: + \: ( {b}^{2} )^{2} \: - \: 2 {a}^{2} {b}^{2} \} \: + \: \{( {c}^{2} )^{2} \: + \: ( {d}^{2} )^{2} \: - \: 2 {c}^{2} {d}^{2} \} \: = \: - 2 \{(ab)^{2} \: + \: (cd)^{2} \: - \: 2abcd \} \\ \\ \\ \sf \implies( {a}^{2} \: - \: {b}^{2} )^{2} \: + \: ( {c}^{2} \: - \: {d}^{2} )^{2} \: = \: - 2(ab \: - \: cd)^{2} \\ \\ \\ \sf \implies( {a}^{2} \: - \: {b}^{2} )^{2} \: + \: ( {c}^{2} \: - \: {d}^{2} )^{2} \: + \: 2(ab \: - \: cd)^{2} \: = \: 0$




$\textsf{We know that square of any integer is always positive ,} \\ \textsf{so if above equation is true then : } \\ \\ \sf \implies (a^2 \: - \: b^2)^2 \: = \: 0 \\ \\ \sf \implies a^2 \: - \: b^2 \: = \: 0 \\ \\ \sf \implies a^2 \: = \: b^2 \\ \\ \sf \: \: \therefore \: \: a \: = \: b \qquad...(1)\\ \\ \textsf{And,} \\ \\ \sf \implies (c^2 \: - \: d^2)^2 \: = \: 0 \\ \\ \sf \implies c^2 \: - \: d^2 \: = \: 0 \\ \\ \sf \implies c^2 \: = \: d^2 \\ \\ \sf \: \: \therefore \: \: c \: = \: d \qquad...(2) \\ \\ \textsf{Again,} \\ \\ \sf \implies2(ab \: - \: cd) \: = \: 0 \\ \\ \sf \implies ab \: - \: cd \: = \: 0$



$\textsf{Plug the value of (1) and (2), } \\ \\ \sf \implies b^2 \: - \: d^2 \: = \: 0 \\ \\ \sf \implies b^2 \: = \: d^2 \\ \\ \sf \: \: \therefore \: \: b \: = \: d \qquad...(3) \\ \\ \textsf{From (1) , (2) and (3) we get : } \\ \\ \sf \implies a \: = \: b \: = \: c \: = \: d$




$\underline{\underline{\textsf{\Large{Proved !! }}}}$