(i) a(b2 + c2) = c(a2 + b2)
(ii) a2b2c2 [1/a3 + 1/b3 + 1/c3] = a3 + b3 + c3
Solution:
(i) a(b2 + c2) = c(a2 + b2)
According to the question, a, b, c are in GP.
Making use of the property of geometric mean, we can write:
b2 = ac
Let us take the LHS: a(b2 + c2)
Upon putting b2 = ac in above equation, we get
$ =a\left( ac\text{ }+\text{ }{{c}^{2}} \right) $
$ ={{a}^{2}}c\text{ }+\text{ }a{{c}^{2}} $
$ =c\left( {{a}^{2}}~+\text{ }ac \right) $
Substitute ac = b2 we get,
$ c\left( {{a}^{2}}~+\text{ }{{b}^{2}} \right)\text{ }=\text{ }RHS $
∴ LHS = RHS
Hence proved.
(ii) a2b2c2 [1/a3 + 1/b3 + 1/c3] = a3 + b3 + c3
According to the question, a, b, c are in GP.
Making use of the property of geometric mean, we can write:
b2 = ac
Let us take the LHS: a2b2c2 [1/a3 + 1/b3 + 1/c3]
$ ={{a}^{2}}{{b}^{2}}{{c}^{2}}/{{a}^{3}}~+\text{ }{{a}^{2}}{{b}^{2}}{{c}^{2}}/{{b}^{3}}~+\text{ }{{a}^{2}}{{b}^{2}}{{c}^{2}}/{{c}^{3}} $
$ ={{b}^{2}}{{c}^{2}}/a\text{ }+\text{ }{{a}^{2}}{{c}^{2}}/b\text{ }+\text{ }{{a}^{2}}{{b}^{2}}/c $
$ =\left( ac \right){{c}^{2}}/a\text{ }+\text{ }{{\left( {{b}^{2}} \right)}^{2}}/b\text{ }+\text{ }{{a}^{2}}\left( ac \right)/c\text{ }\left[ by\text{ }substituting\text{ }the\text{ }{{b}^{2}}~=\text{ }ac \right] $
$ =a{{c}^{3}}/a\text{ }+\text{ }{{b}^{4}}/b\text{ }+\text{ }{{a}^{3}}c/c $
$ ={{c}^{3}}~+\text{ }{{b}^{3}}~+\text{ }{{a}^{3}}~=\text{ }RHS $
∴ LHS = RHS
Hence proved.