If a - b = 6 and a^2 +b^2= 20, find the value of (1) ab
(ii) a^3 - b^3
Answers
Answered by
0
Answer:
ab= -8
a^3 - b^3=72
Explanation:
(1)(a-b)^2=(a^2+b^2)-2ab
(6)^2=20-2ab
36=20-2ab
2ab=20-36
ab= -16/2
ab= -8
(ii)a^3 - b^3=(a-b)[a^2+b^2+ab]
=6(20-8)
=6×12
a^3 - b^3=72
Answered by
0
Answer:
Given , a+b = 6 ( equation 1 )
a²+b² = 20 ( equation 2 )
Identify: (a +b)² = a² + b² + 2ab
Putting values of equation 1 and 2 here
( 6 )² = 20 + 2ab
36 - 20 = 2ab
16 = 2ab
8 = ab Or
ab = 8 ( equation 3 )
Now identity, a³+ b³ = (a+b)(a² + b² - ab)
Putting equation 1 , 2 and 3 here
a³ + b³ = (6) (20 - 8)
a³ + b³ = 6 × 12
a³+ b³ = 72
Explanation:
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