We have, a+b+c=1,ab+bc+ca=2 and abc=3Now, a2+b2+c2=(a+b+c)2−2Σab=−3
⇒(ab+bc+ca)2=Σ(ab)2+2abcΣa
⇒Σ(ab)2=−2
So, a4+b4+c4=(a2+b2+c2)2−2Σ(ab)2
=9−2(−2)
=13.
JEE Main 2021 — Mathematics Algebra
If a+b+c=1,ab+bc+ca=2 and abc=3, then the value of a4+b4+c4 is equal to:
Held on 25 Jul 2021 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
Sign in to keep a private note on this question. Nothing you write is ever public.
The number of values of $z \in \mathbb{C}$, satisfying the equations $|z-(4+8i)|=\sqrt{10}$ and $|z-(3+5i)|+|z-(5+11i)|=4\sqrt{5}$, is:
Let $a_1, a_2, a_3, \ldots$ be an A.P. and $g_1 = a_1, g_2, g_3, \ldots$ be an increasing G.P. If $a_1 = a_2 + g_2 = 1$ and $a_3 + g_3 = 4$, then $a_{10} + g_5$ is equal to:
Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^{k}}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to
Let $A=\{z \in \mathbb{C}:|z-2| \leqslant 4\}$ and $B=\{z \in \mathbb{C}:|z-2|+|z+2|=5\}$. Then the max $\left\{\left|z_{1}-z_{2}\right|: z_{1} \in \mathrm{~A}\right.$ and $\left.z_{2} \in \mathrm{~B}\right\}$ is :
A building has ground floor and 10 more floors. Nine persons enter in a lift at the ground floor. The lift goes up to the $10^{\text{th}}$ floor. The number of ways, in which any 4 persons exit at a floor and the remaining 5 persons exit at a different floor, if the lift does not stop at the first and the second floors, is equal to :
Work through every JEE Main Algebra PYQ, year by year.