Question

if 'b' is the geometric mean of 'a' and 'c ' then show that
a²b²c²[1/a³ +1/b³+1/c³] =a³+b³+c³​

Answer 1

Answer:

Solution

a:b = b:c

a\: =\: \dfrac{b^2}{c}a=cb2

Let take LHS of Equation

a^2b^2c^2\Big(\dfrac{1}{a^3}\: +\: \dfrac{1}{b^3}\: +\: \dfrac{1}{c^3}\Big)a2b2c2(a31+b31+c31)

Let we put value of a in equation after simplification

\dfrac{a^2b^2c^2}{a^3}\: +\: \dfrac{a^2b^2c^2}{b^3}\: +\: \dfrac{a^2b^2c^2}{c^3}a3a2b2c2+b3a2b2c2+c3a2b2c2

\dfrac{b^2c^2}{a}\: +\: \dfrac{a^2c^2}{b}\: +\: \dfrac{a^2b^2}{c}ab2c2+ba2c2+ca2b2

\dfrac{b^2c^2}{\dfrac{b^2}{c}}\: +\: \dfrac{\Big(\dfrac{b^2}{c}\Big)^2c^2}{b}\: +\: a^2.acb2b2c2+b(cb2)2c2+a2.a

\dfrac{b^2c^2.c}{b^2}\: +\: \dfrac{\dfrac{b^4}{c^2}c^2}{b}\: +\: a^2.ab2b2c2.c+bc2b4c2+a2.a

c^2.c \:+\: \dfrac{b^4}{c^2}\times\dfrac{c^2}{b}+\: a^2.ac2.c+c2b4×bc2+a2.a

c^3 \:+\: b^3+\: a^3c3+b3+a3

Rearrange the sequence

\boxed{a^3 \:+\: b^3+\: c^3}a3+b3+c3

RHS

LHS = RHS

Hence Proved

Step-by-step explanation:

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Answer 2

Answer:

Here you go

Step-by-step explanation:

All the best