Question

If $a,b,c$ are in $A.P$., prove that $a^3+4b^3+c^3=3b(a^2+c^2)$.

Answer 1

$a,b,c$ are in A.P

$2b=a+c\Rightarrow b=\frac{a+c}{2}$

Consider L.H.S

$a^3+4b^3+c^3$

$=a^3+c^3+4\times \frac{(a+c{)}^3}{8}$

$=a^3+c^3+\frac{1}{2}(a+c{)}^3$

$=\frac{3}{2}(a+c{)}^3-3ac(a+c)$ $[\because a^3+c^3=(a+c{)}^3-3ac(a+c)]$

$=\frac{3}{2}(a+c)[a^2+c^2+2ab-2ab]$

$=\frac{3}{2}(a+c)(a^2+c^2)$

$=3\left(\frac{a+c}{2}\right)(a^2+c^2)$

$=3b(a^2+c^2)=R.H.S$

Hence proved.