$a^{2} b^{2} c^{2} (a^{- 3} + b^{- 3} + c^{- 3}) = (a^{3} + b^{3} + c^{3})$

Open in App

Solution

$\frac{a}{b} = \frac{b}{c} = k$
$a = b k$
$b = c k$
$a = c k^{2}$
$\frac{a + b + c}{a - b + c} = \frac{c k^{2} + c k + c}{c k^{2} - c k + c}$
$= \frac{c (k^{2} + k + 1)}{c (k^{2} - k + 1)}$
Now,
$\frac{{(a + b + c)}^{2}}{(a^{2} + b^{2} + c^{2})} = \frac{{(c k^{2} + c k + c)}^{2}}{c^{2} k^{4} + c^{2} k^{2} + c^{2}}$
$= \frac{c^{2} {(k^{2} + k + 1)}^{2}}{c^{2} (k^{4} + k^{2} + 1)}$
$= \frac{(k^{2} + k + 1) (k^{2} + k + 1)}{(k^{2} + k + 1) (k^{2} - k + 1)}$
$= \frac{k^{2} + k + 1}{k^{2} - k + 1}$
factors of $k^{4} + k^{2} + 1 = (k^{2} + k + 1) (k^{2} - k + 1)$

Suggest Corrections

Similar questions

a + b + c = 1; a^{2} + b^{2} + c^{2} = 2; a^{3} + b^{3} + c^{3} = 3

3 [\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - 1^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}}]

a + b + c = 3

a^{2} + b^{2} + c^{2} = 6

a^{3} + b^{3} + c^{3} = 8

a^{4} + b^{4} + c^{4} = ?

If a + b + c = 9, and $a^{2} + b^{2} + c^{2} = 35$ , find the value of $a^{3} + b^{3} + c^{3} - 3 a b c .$

\frac{(a^{2} - b^{2}) + {(b^{2} - c^{2})}^{3} + {(c^{2} - a^{2})}^{3}}{{(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}}

Join BYJU'S Learning Program