Question

if a,b,c are in A.P then prove that a3+4b3+c3= 3(a2+c2),

Answer 1

a, b and c are in A.P
so , common difference always constant.
b - a = c - b
b = (a + c)/2
2b = (a + c)

now,

LHS = a³ + 4b³ + c³
= a³ + c³ + 1/2(2b)³
= a³ + c³ + 1/2 (a + c)³
=3/2 (a + c)³ - 3ac(a + c)
= 3/2( a + c) { a² + c² + 2ab - 2ab }
= 3/2(a + c)(a² + c²)
= 3{(a + c)/2}( a² + c²)
= 3b (a² + c²)

hence,
a³ + 4b³ + c³ = 3b(a² + c²)

Answer 2

Shortcut method

A, b, c are in ap

Let a=1 b=2 c=3

Lhs= a³+4b³+c³

=1+4x8+27

=60

Rhs=3b(a²+c²)

=3x2(1+9)

=60

Hence,

a³+4b³+c³=3b(a²+c²)