Question

If a, b, c are the zeroes of x^3 – 3x + 1 = 0, then find the value of [(a - 2)(b − 2)(c - 2)]^2?

Please provide a step-by-step solution.​

Answer 1

f(x) =

let a,b and c are zeroes of f(x) = x³-3x+1 = 0

sum of zeroes

a + b + c = - (cofficient of x²)/(cofficient of x³)

a+b+c = - 0/1 = 0

sum of zeroes two at a time

ab + bc + ca = cofficient of x /cofficient of x³

ab + bc + ca = -3

product of zeroes

a×b×c = - constant term/ cofficient of x³

a×b×c = -1

[(a - 2)(b − 2)(c - 2)]²

[( ab - 2a - 2b + 4)(c-2)]²

[(c-2)]²[ c( ab - 2a - 2b + 4)-2(ab - 2a - 2b + 4)]²

(abc - 2ac - 2bc + 4c - 2ab + 4a + 4b - 8)²

(abc -2ab - 2bc - 2ca + 4a + 4b + 4c -8)²

(abc - 2( ab + bc +ac ) +4( a + b + c ) -8)²

(-1 -2 × -3 + 4 × 0 - 8 )²

(-1 + 6 + 0 - 8)²

(-9 + 6)²

(-3)²

9 ans

Answer 2

$\large\bf\underline \orange{Given:-}$

• a , b and c are Zeroes of polynomial x³-3x+1 = 0

$\large\bf\underline \orange{To \: find:-}$

• value of[(a - 2)(b − 2)(c - 2)]^2

$\huge\bf\underline \green{Solution:-}$

Let α , β and γ be the zeroes of the given polynomial.

• Let α = a
• β = b
• γ = c

p(x) = x³-3x+1 = 0

• a = 1
• b = 0
• c = -3
• d = 1

$\pink{\bf \alpha + \beta + \gamma = \frac{ - b}{a} }$

$: \implies \rm \: a + b + c = \frac{0}{1} \\ \\ : \implies \bf \: a + b + c = 0$

$\pink{ \bf \: \alpha \beta + \beta \gamma + \gamma \alpha = \frac{c}{a} }$

$: \implies \rm \: ab + bc + ca = \frac{ - 3}{1} \\ \\ : \implies \bf \: ab + bc + ca = - 3$

$\bf \pink{ \alpha \beta \gamma = \frac{ - d}{a} }$

$: \implies \rm \: abc = \frac{ - 1}{1} \\ \\ : \implies \bf \: abc = - 1$

Now,

$\mapsto \rm \: [(a - 2)(b - 2)(c - 2)] {}^{2} \\ \\ \mapsto \rm \: [a(b - 2) - 2(b - 2)(c - 2)] {}^{2} \\ \\ \mapsto \rm \:[ab - 2a - 2b + 4(c - 2)] {}^{2} \\ \\ \mapsto \rm \small[c(ab - 2a - 2b + 4) - 2(ab - 2a - 2b + 4)] {}^{2} \\ \\ \mapsto \rm \small [abc - 2ac - 2bc + 4c - 2ab + 4a + 4b - 8] {}^{2} \\ \\ \mapsto \rm \small[abc - 2(ac + bc + ca) + 4(a + b + c) - 8] {}^{2} \\ \\ \mapsto \rm[ - 1 - 2( - 3) + 4(0) - 8] {}^{2} \\ \\ \mapsto \rm[ - 1 + 6 +0 - 8] {}^{2} \\ \\ \mapsto \rm[ 5 - 8] {}^{2} \\ \\ \mapsto \rm[ - 3] {}^{2} \\ \\ \bf \red{ = 9}$

Hence,

Value of $\rm \: [(a - 2)(b - 2)(c - 2)] {}^{2}$ = 9