If a+b+c=1,ab+bc+ca=2 and abc=3,then the valu | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:$1^{2}

Vedclass · Accepted Answer

$13$

Vedclass · Answer

(B) Given $a+b+c=1$,$ab+bc+ca=2$,and $abc=3$.First,find $a^{2}+b^{2}+c^{2}$ using the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2} + 2(ab+bc+ca)$:$1^{2} = a^{2}+b^{2}+c^{2} + 2(2)$$a^{2}+b^{2}+c^{2} = 1 - 4 = -3$.Next,find $a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2}$ using the identity $(ab+bc+ca)^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2abc(a+b+c)$:$2^{2} = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 2(3)(1)$$4 = a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} + 6$$a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2} = 4 - 6 = -2$.Finally,use the identity $a^{4}+b^{4}+c^{4} = (a^{2}+b^{2}+c^{2})^{2} - 2(a^{2}b^{2}+b^{2}c^{2}+c^{2}a^{2})$:$a^{4}+b^{4}+c^{4} = (-3)^{2} - 2(-2)$$a^{4}+b^{4}+c^{4} = 9 + 4 = 13$.

If $a+b+c=1$,$ab+bc+ca=2$ and $abc=3$,then the value of $a^{4}+b^{4}+c^{4}$ is equal to $...$

Similar Questions

The values of $p$ for which one root of the equation $x^2 - 30x + p = 0$ is the square of the other are:

Let $\alpha, \beta$ be the roots of the equation $ax^2 + 2bx + c = 0$ and $\gamma, \delta$ be the roots of the equation $px^2 + 2qx + r = 0$. If $\alpha, \beta, \gamma, \delta$ are in $G.P.$,then:

If one root of the equation $ax^2 + bx + c = 0$ is the square of the other root,then $b^3 + a^2c + ac^2 = \dots$

If $\alpha, \beta, \gamma$ are the roots of the equation $x^3-3x^2+3x+1=0$,then $\alpha^2\beta^2+\beta^2\gamma^2+\gamma^2\alpha^2=$

The equation $x^5-5x^3+5x^2-1=0$ has three equal roots. If $\alpha$ and $\beta$ are the other two roots of this equation,then $\alpha+\beta+\alpha\beta=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module