Not an inequality

Algebra Level 2

Let a a , b b , and c c be three real numbers such that a + b + c = − 1 a + b + c = -1 and a b c = 1 abc = 1 . Find the value of the expression:

a 4 − 3 a + b 4 − 3 b + c 4 − 3 c \frac{a^4 - 3}a + \frac{b^4 - 3}b + \frac{c^4 - 3}c


The answer is 2.

Shikhar Srivastava by Shikhar Srivastava , 22, India

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2 solutions

We know that for any real numbers a , b a\,,\,b\, and c \,c ,

a 3 + b 3 + c 3 − 3 a b c = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) ⇒ a 3 + b 3 + c 3 − 3 ( 1 ) = ( − 1 ) ( a 2 + b 2 + c 2 − a b − b c − c a ) ⇒ a 3 + b 3 + c 3 − 3 = − a 2 − b 2 − c 2 + a b + b c + c a ⇒ a 3 + b 3 + c 3 − 3 a b − 3 b c − 3 c a = 3 − a 2 + b 2 + c 2 − 2 a b − 2 b c − 2 c a ⇒ a 3 + b 3 + c 3 − 3 a b − 3 b c − 3 c a = 3 − ( a + b + c ) 2 ⇒ a 3 + b 3 + c 3 − 3 c a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)\newline\Rightarrow a^3 + b^3 + c^3 - 3(1) = (-1)(a^2 + b^2 + c^2 - ab - bc - ca)\newline\Rightarrow a^3 + b^3 + c^3 - 3 = -a^2 - b^2 - c^2 + ab + bc + ca\newline\Rightarrow a^3 + b^3 + c^3 - 3ab - 3bc - 3ca = 3 - a^2 + b^2 + c^2 - 2ab - 2bc - 2ca\newline \Rightarrow a^3 + b^3 + c^3 - 3ab - 3bc - 3ca = 3 - (a + b + c)^2 \newline\Rightarrow a^3 + b^3 + c^3 - \Large\frac{3}{c} − 3 a - \Large\frac{3}{a} − 3 b - \Large\frac{3}{b} = 3 − ( − 1 ) 2 = 2 = 3 - (-1)^2 = 2

⇒ a 4 − 3 a \Rightarrow \Large\frac{a^4 - 3}{a} + b 4 − 3 b + \Large\frac{b^4 - 3}{b} + c 4 − 3 c + \Large\frac{c^4 - 3}{c} = 2 = 2

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No need to add space and separate the formula with \ ( \ ) just use \ [ and \ ]. I have amended your problem. I wonder if you can still edit it to see. If not look at below. This was I have done. LaTex is much easier to use. Of course we have to assume its creator are smart.

Chew-Seong Cheong - 1 year, 1 month ago
Chew-Seong Cheong
Apr 17, 2020

X = a 4 − 3 a + b 4 − 3 b + c 4 − 3 c = a 3 + b 3 + c 3 − 3 ( 1 a + 1 b + 1 c ) = ( a + b + c ) ( a 2 + b 2 + c 2 − a b − b c − c a ) + 3 a b c − 3 ( 1 a + 1 b + 1 c ) = ( a + b + c ) ( ( a + b + c ) 2 − 3 ( a b + b c + c a ) ) + 3 a b c − 3 ( 1 a + 1 b + 1 c ) = ( a + b + c ) ( ( a + b + c ) 2 − 3 a b c ( 1 a + 1 b + 1 c ) ) + 3 a b c − 3 ( 1 a + 1 b + 1 c ) = ( − 1 ) ( ( − 1 ) 2 − 3 ( 1 a + 1 b + 1 c ) ) + 3 − 3 ( 1 a + 1 b + 1 c ) = 2 \begin{aligned} X & = \frac {a^4-3}a + \frac {b^4-3}b + \frac {c^4-3}c \\ & = a^3 + b^3 + c^3 - 3\left(\frac 1a+\frac 1b+\frac 1c\right) \\ & = (a+b+c)\left(a^2+b^2+c^2 - ab-bc-ca\right) + 3abc - 3\left(\frac 1a+\frac 1b+\frac 1c\right) \\ & = (a+b+c)\left((a+b+c)^2 -3(ab+bc+ca) \right) + 3abc - 3\left(\frac 1a+\frac 1b+\frac 1c\right) \\ & = (a+b+c)\left((a+b+c)^2 -3abc\left(\frac 1a+\frac 1b+\frac 1c\right) \right) + 3abc - 3\left(\frac 1a+\frac 1b+\frac 1c\right) \\ & = (-1)\left((-1)^2 -3\left(\frac 1a+\frac 1b+\frac 1c \right) \right) + 3 - 3\left(\frac 1a+\frac 1b+\frac 1c\right) \\ & = \boxed 2 \end{aligned}

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