From a+b=c+d

=> (a+b)²=(c+d)²

=>ab=cd

Consider (a+b)³=(c+d)³

We have a³+3ab(a+b)+b³=c³+3cd+d³

=>a³+b³=c³+d³

Consider (a³+b³).(a+b)=(c³+d³).(c+d)

=> a⁴+ab(a²+b²)+b⁴=c⁴+cd(c²+d²)

=>a⁴+b⁴=c⁴+d⁴

We can infer this

aⁿ+bⁿ=cⁿ+dⁿ

=> (aⁿ+bⁿ).(a+b)=(cⁿ+dⁿ).(c+d)

=> aⁿ⁺¹+ab(a^n-1+b^n-1)+bⁿ⁺¹=cⁿ⁺¹+cd(c^n-1+d^n-1)+dⁿ⁺¹

=>aⁿ⁺¹+bⁿ⁺¹=cⁿ⁺¹+dⁿ⁺¹

So we have a²⁰¹⁸+b²⁰¹⁸=c²⁰¹⁸+d²⁰¹⁸

Given the real numbers a,b and c such that \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=\dfrac{4}{a+b+c}\)

Find the value of the expression

\(M=\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}\)