Basics & LogarithmseasyPYQ · JEE Main · 25 Jul 2021 · Shift 1 (Morning)Free

Basics & Logarithms: Abc Value Equal (JEE Main 2021)

JEE Maths question with a full step-by-step solution.

Question
If a+b+c=1a+b+c=1, ab+bc+ca=2ab+bc+ca=2 and abc=3abc=3, then the value of a4+b4+c4a^{4}+b^{4}+c^{4} is equal to
Solution
Answer: 13
Step 1: Use the identity:
a2+b2+c2=(a+b+c)2−2(ab+bc+ca)=1−2(2)=−3a^{2}+b^{2}+c^{2}=(a+b+c)^{2}-2(ab+bc+ca)=1-2(2)=-3
 Step 2:
(ab+bc+ca)2=∑(ab)2+2∑ab2c=4(ab+bc+ca)^{2}=\sum(ab)^{2}+2\sum ab^{2}c=4
∑(ab)2=(ab+bc+ca)2−2abc(a+b+c)=4−2(3)(1)=−2\sum(ab)^{2}=(ab+bc+ca)^{2}-2abc(a+b+c)=4-2(3)(1)=-2
 Step 3:
a4+b4+c4=(a2+b2+c2)2−2[(ab)2+(bc)2+(ca)2]a^{4}+b^{4}+c^{4}=(a^{2}+b^{2}+c^{2})^{2}-2[(ab)^{2}+(bc)^{2}+(ca)^{2}]
=(−3)2−2(−2)=9+4=13=(-3)^{2}-2(-2)=9+4=13
 Answer: 1313
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Basics & Logarithms · easy
If x=∏n=12026nx = \prod_{n=1}^{2026} n, then the value of the expression 11log⁡2x+1log⁡3x+⋯+1log⁡2026x\frac{1}{\dfrac{1}{\log_2 x} + \dfrac{1}{\log_3 x} + \cdots + \dfrac{1}{\log_{2026} x}} is
Basics & Logarithms · easy
Number of integral values of xx satisfying the inequality (34)6x+10−x2<2764\left(\dfrac{3}{4}\right)^{6x+10-x^2} < \dfrac{27}{64} is:
Basics & Logarithms · easy
Number of natural numbers for which the number log⁡4−x(x2−14x+45)\log_{4-x}(x^2-14x+45) is defined is:
Basics & Logarithms · easy
The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_{e}x)^{2}+3}=x^{8}, x>0x>0, is

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath