Join 60,000+ competitive exam aspirants
or with email
Your Cart
Subtotal₹0
Total₹0
Examoogle • User • info@examoogle.com • EE-2024-8821
Chapter 1 of 12 • Page 1 of 248🔒 Protected PDF • Watermarked
Join 60,000+ competitive exam aspirants
If a−b=1 and a3−b3=61, then the value of ab is:
20
60
20
30
20
Use the identity a3−b3=(a−b)((a−b)2+3ab). Given 61=1×(12+3ab), which simplifies to 61=1+3ab, so 60=3ab, and ab=20.
a - b = 1 and a³ - b³ = 61
(a−b)3=a3−b3−3ab(a−b)
Use the identity a3−b3=(a−b)((a−b)2+3ab). Given 61=1×(12+3ab), which simplifies to 61=1+3ab, so 60=3ab, and ab=20.
Confusing the sign of the 3ab term in the expansion of (a−b)3 or misapplying the formula for a3−b3.
Use algebraic identity
We use the identity relating the difference of cubes to the cube of the difference: a3−b3=(a−b)3+3ab(a−b).
a3−b3=(a−b)3+3ab(a−b)
Substitute given values
Substitute a−b=1 and a3−b3=61 into the identity.
61=(1)3+3ab(1)
Solve for ab
Simplify the equation: 61=1+3ab, which leads to 3ab=60.
ab=360=20
A is correct because substituting the given values into the algebraic identity a3−b3=(a−b)3+3ab(a−b) yields ab=20.
This type of problem frequently appears in competitive exams. Remember the two key identities: a3−b3=(a−b)(a2+ab+b2) and a3−b3=(a−b)3+3ab(a−b).
Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.
Free Download • 100K+ Aspirants