Question

If a,b,c are such that a + b + c = 2. a2 + b2 + b2 = 6, a3 + b3 + c3 = 8, then a4 + b4 +c4 is equal to
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Answer 1

Step-by-step explanation:

The following method can solve for [math]a^k+b^k+c^k[/math] .

[math]ab+bc+ca = \frac{(a+b+c)^2 - (a^2+b^2+c^2)}{2}[/math]

[math]ab+bc+ca = \frac{16–6}{2}[/math]

[math]ab+bc+ca = 5[/math]

[math]abc = \frac{(a^3+b^3+c^3)-(a+b+c)(a^2+b^2+c^2-(ab+bc+ca))}{3}[/math]

[math]abc = \frac{4}{3}[/math]

Construct an equation with a,b,c as roots

[math]x^3 - (a+b+c)x^2 + (ab+bc+ca)x - abc = 0[/math]

[math]x^3 - 4x^2 + 5x -\frac{4}{3} = 0[/math]

[math]3x^3 - 12x^2 + 15x - 4= 0[/math]

Using this construct an equation with [math]a^4,b^4,c^4[/math] as roots.

This can be done by replace [math]x[/math] with [math]x^{\frac{1}{4}}[/math] and convert it into a cubic equation

Let [math]y = x^{\frac{1}{4}}[/math]

[math]3y^3 - 12y^2 + 15y - 4= 0[/math]

[math]3y^3 + 15y = 12y^2+4[/math]

squaring on both sides

[math]9y^6+90y^4+225y^2 = 144y^4 + 96y^2+16[/math]

[math]9y^6+129y^2 = 54y^4+16[/math]

squaring on both sides

[math]81y^{12}+2322y^8+16641y^4 = 2916y^8+1728y^4+256[/math]

replace [math]y^4[/math] with [math]x[/math]

[math]81x^3+2322x^2+16641x = 2916x^2+1728x+256[/math]

[math]81x^3 - 594x^2 + 14913x - 256 = 0[/math]

Now sum of roots = [math]\frac{594}{81}[/math]

[math]a^4+b^4+c^4 = \frac{22}{3}[/math]

[math][a^4+b^4+c^4] = 7[/math]

HOPE THIS HELPS YOU

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