Prove-that-in-ABC-a-3-cos-B-C-b-3-cos-C-A-c-3-cos-A-B-3abc-

Question Number 15888 by Tinkutara last updated on 15/Jun/17

Prove that in Δ A B C, a^{3} \cos (B - C) +

b^{3} \cos (C - A) + c^{3} \cos (A - B) = 3 a b c

Answered by ajfour last updated on 15/Jun/17

Commented by ajfour last updated on 15/Jun/17

I n Δ A B C J o i n B t o D s u c b t h a t

∠ A B D = ∠ C

t h e n Δ A D B \sim Δ A B C

\cos (B - C) = \frac{{k a}^{2} + a^{2} - {(b - k c)}^{2}}{2 {k a}^{2}}

s i d e A B = c = k b

s o k = c / b

\cos (B - C) = \frac{\frac{a^{2} c^{2}}{b^{2}} + a^{2} - {(b - \frac{c^{2}}{b})}^{2}}{\frac{2 a^{2} c}{b}}

\cos (B - C) = \frac{a^{2} c^{2} + a^{2} b^{2} - {(b^{2} - c^{2})}^{2}}{2 a^{2} b c}

a^{3} \cos (B - C) = \frac{a^{2} [a^{2} c^{2} + a^{2} b^{2} - {(b^{2} - c^{2})}^{2}]}{2 a b c}

s i m i l a r l y

b^{3} \cos (C - A) = \frac{b^{2} [b^{2} a^{2} + b^{2} c^{2} - {(c^{2} - a^{2})}^{2}]}{2 a b c}

c^{3} \cos (A - B) = \frac{c^{2} [c^{2} b^{2} + c^{2} a^{2} - {(a^{2} - b^{2})}^{2}]}{2 a b c}

s u m m i n g t h e m u p, w e g e t

Σ a^{3} \cos (B - C) = \frac{6 a^{2} b^{2} c^{2}}{2 a b c} = 3 a b c .

Commented by Tinkutara last updated on 15/Jun/17

Thanks Sir!