find-the-value-of-a-b-and-c-a-b-c-4-a-2-b-2-c-2-66-a-3-b-3-c-3-280-

Question Number 12261 by frank ntulah last updated on 17/Apr/17

f i n d t h e v a l u e o f a b a n d c

a + b + c = 4

a^{2} + b^{2} + c^{2} = 66

a^{3} + b^{3} + c^{3} = 280

Answered by mrW1 last updated on 17/Apr/17

{(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = 4^{2} = 16

\Rightarrow a b + b c + c a = \frac{16 - 66}{2} = - 25

{(a + b + c)}^{3} = a^{3} + b^{3} + c^{3} + 6 a b c + 3 (a^{2} b + a^{2} c + b^{2} a + b^{2} c + c^{2} a + c^{2} b)

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

= 280 + 3 \times 4 \times (- 25) - 3 a b c = 4^{3} = 64

\Rightarrow a b c = \frac{280 + 3 \times 4 \times (- 25) - 64}{3} = - 28

a, b, c a r e t h e r o o t s o f f o l l o w i n g c u b i c e q n .

x^{3} - (a + b + c) x^{2} + (a b + b c + c a) x - a b c = 0

\Rightarrow x^{3} - 4 x^{2} - 25 x + 28 = 0

\Rightarrow (x - 1) (x + 4) (x - 7) = 0

\Rightarrow x_{1} = - 4, x_{2} = 1, x_{3} = 7

p o s s i b l e s o l u t i o n s f o r a, b, c a r e :

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} - 4 \\ 1 \\ 7 \end{matrix})

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} - 4 \\ 7 \\ 1 \end{matrix})

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} 1 \\ - 4 \\ 7 \end{matrix})

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} 1 \\ 7 \\ - 4 \end{matrix})

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} 7 \\ - 4 \\ 1 \end{matrix})

(\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} 7 \\ 1 \\ - 4 \end{matrix})

Commented by frank ntulah last updated on 17/Apr/17

t h a n k y o u s i r y o u m a k e m e h a p p y

Commented by tawa last updated on 17/Apr/17

Yes sir, check line 4 \dots \dots a^{3} + b^{3} + c^{3} \dots . .

Commented by mrW1 last updated on 17/Apr/17

T h a n k s . Y o u a r e r i g h t . I h a v e c o r r e c t e d .

Commented by tawa last updated on 17/Apr/17

Yes sir . Thank you . i always follow your steps very well . God bless you sir .

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