UPSC CDS 1 2022 Math Paper

Given:
a+b=2,‌

1

a

+‌

1

b

=2
Formual Used:
a3+b3=(a+b)(a2+b2−ab)
Calculation:
We have a+b=2‌‌⋯−(i)
And ‌

1

a

+‌

1

b

=2
⇒a+b=2ab
⇒2=2ab‌‌[‌ from (i) ‌]
⇒ab=1‌‌⋯‌ (ii) ‌
Squaring the equation (i), we get
a2+b2+2ab=4
⇒a2+b2+2×1=4
⇒a2+b2=2
According to the formula used
a3+b3=(a+b)(a2+b2−ab)
And from equation (i), (ii) and (iii), we get
⇒a3+b3=2(2−1)
⇒a3+b3=2
∴ The required value of a3+b3 is 2 .