Problem No 52

Algebra Level 3

Suppose a , b a,b and c c are the roots of x 3 − 4 x + 1 = 0 x^3 - 4x + 1 = 0 , find the value of a 2 b c a 3 + 1 + a b 2 c b 3 + 1 + a b c 2 c 3 + 1 . \dfrac{a^2 bc}{a^3+1} + \dfrac{ab^2 c}{b^3+1} + \dfrac{abc^2}{c^3+1}.

3 4 \frac{3}{4} 4 3 \frac{4}{3} − 3 4 -\frac{3}{4} − 4 3 -\frac{4}{3}

View wiki
Benedict Dimacutac by Benedict Dimacutac , 21, Philippines

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Let a , b , c a,b,c be the roots of x 3 − 4 x + 1 = 0 x^3 - 4x + 1 =0

⇒ a 3 + 1 = 4 a , b 3 + 1 = 4 b , c 3 + 1 = 4 c \Rightarrow a^3 + 1 = 4a , b^3 + 1 = 4b , c^3 + 1 = 4c

⇒ S = a 2 b c a 3 + 1 + a b 2 c b 3 + 1 + a b c 2 c 3 + 1 = a b c ( a a 3 + 1 + b b 3 + 1 + c c 3 + 1 ) \Rightarrow S = \dfrac{a^2bc}{a^3 + 1} + \dfrac{ab^2c}{b^3 + 1} + \dfrac{abc^2}{c^3 + 1} = abc\left(\dfrac{a}{a^3 + 1} + \dfrac{b}{b^3 + 1} + \dfrac{c}{c^3 + 1} \right)

⇒ S = ( − 1 ) ( a 4 a + b 4 b + c 4 c ) = − 3 4 \Rightarrow S = (-1)\left(\dfrac{a}{4a} + \dfrac{b}{4b} + \dfrac{c}{4c} \right) = \boxed{-\dfrac{3}{4}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

But you have to show ,How you got ⇒ a 3 + 1 = 4 a \Rightarrow a^3 + 1 = 4a

and others.

Archiet Dev - 5 years, 6 months ago

Log in to reply

Let f ( x ) = x 3 − 4 x + 1 f(x) = x^3 - 4x + 1 . Then since a,b, & c are the roots,

f ( a ) = f ( b ) = f ( c ) = 0 f(a) = f(b) = f(c) = 0

a 3 − 4 a + 1 = b 3 − 4 b + 1 = c 3 − 4 c + 1 = 0 a^3 - 4a + 1 = b^3 - 4b + 1 = c^3 -4c + 1 = 0

Hope you understood.

Vishwak Srinivasan - 5 years, 6 months ago

Log in to reply

Yeah,I got it... Thanks

Archiet Dev - 5 years, 6 months ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...