a,b,c are in continued propertion prove (a3+b3+c3)/a2b2c2=1/a3 + 1/b3+ 1/c3 - Maths -

Question

a,b,c are in continued propertion prove

(a3+b3+c3)/a2b2c2=1/a3
+ 1/b3+ 1/c3

Aakash Sharma · Accepted Answer

Given : a, b, c are in continued proportion $\Rightarrow \frac{a}{b} = \frac{b}{c} or b^{2} = a c \Rightarrow (b^{2})^{2} = (a c)^{2} \Rightarrow b^{4} = a^{2} c^{2}$ Now $\frac{a^{3} + b^{3} + c^{3}}{a^{2} b^{2} c^{2}} = \frac{a^{3}}{a^{2} b^{2} c^{2}} + \frac{b^{3}}{a^{2} b^{2} c^{2}} + \frac{c^{3}}{a^{2} b^{2} c^{2}} = \frac{a}{b^{2} c^{2}} + \frac{b}{a^{2} c^{2}} + \frac{c}{a^{2} b^{2}} = \frac{a}{b} \times \frac{1}{b c^{2}} + \frac{b}{b^{4}} + \frac{c}{b} \times \frac{1}{a^{2} b} = \frac{b}{c} \times \frac{1}{b c^{2}} + \frac{1}{b^{3}} + \frac{b}{a} \times \frac{1}{a^{2} b} = \frac{1}{c^{3}} + \frac{1}{b^{3}} + \frac{1}{a^{3}} = \frac{1}{a^{3}} + \frac{1}{b^{3}} + \frac{1}{c^{3}}$

a,b,c are in continued propertion. prove (a3+b3+c3)/a2b2c2=1/a3 + 1/b3+ 1/c3

a,b,c are in continued propertion. prove

(a³+b³+c³)/a²b²c²=1/a³ + 1/b³+ 1/c³