I-f-a-2-5-b-2-4c-2-4b-a-c-then-find-the-value-of-E-b-c-a-3-abc-abc-0-

Question Number 181593 by mnjuly1970 last updated on 27/Nov/22

I f, a^{2} + 5 b^{2} + 4 c^{2} = 4 b (a + c)

then, find the value of :

E = \frac{{(b + c - a)}^{3}}{a b c} = ? (a b c \neq 0)

Answered by mahdipoor last updated on 27/Nov/22

\Rightarrow a^{2} + 5 b^{2} + 4 c^{2} - 4 b a - 4 b c = 0

\Rightarrow (a^{2} + 4 b^{2} - 4 b a) + (b^{2} + 4 c^{2} - 4 b c) = 0 \Rightarrow

{(a - 2 b)}^{2} + {(b - 2 c)}^{2} = 0 \Rightarrow {\begin{cases} a - 2 b = 0 \Rightarrow a = 2 b \\ b - 2 c = 0 \Rightarrow b = 2 c \end{cases}

E = \frac{{(b + c - a)}^{3}}{a b c} = \frac{{(2 c + c - 4 c)}^{3}}{(4 c) (2 c) (c)} = \frac{- c^{3}}{8 c^{3}} = \frac{1}{8}

Commented by mnjuly1970 last updated on 27/Nov/22

m a m n o o n j e n a b m a h d i p o o r

z e n d e h b a s h i d .

Answered by madhabnandi last updated on 27/Nov/22

A n s :

a^{2} + 5 b^{2} + 4 c^{2} = 4 b (a + c)

\Rightarrow a^{2} + 4 b^{2} + b^{2} + 4 c^{2} = 4 a b + 4 b c

\Rightarrow a^{2} - 4 a b + 4 b^{2} + b^{2} - 4 b c + 4 c^{2} = 0

\Rightarrow {(a - 2 b)}^{2} + {(b - 2 c)}^{2} = 0

∴ a - 2 b = 0 a n d b - 2 c = 0

∴ a = 2 b \Rightarrow b = 2 c .

= 4 c .

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

= \frac{{(2 c + c - 4 c)}^{3}}{4 c \times 2 c \times c}

= \frac{{(- c)}^{3}}{8 c^{3}}

= \frac{- c^{3}}{8 c^{3}}

= - \frac{1}{8} .

Commented by mnjuly1970 last updated on 27/Nov/22

g r a t e f u l s i r