Question

if a+b=10 and a2+b2=58 (square) find a3+b3(cube)

Answer 1

(a+b)3=a3+b3+3ab(a+b)=(10)3------------------------(1)
(a+b)2=a2+b2+2ab=100----------------------(2)
58+2ab=100
2ab=42
ab=21---------------------(4)
from 1 eqn
a3+b3+3(21) (10)=1000-------------------------(From eqn 1 and 4)
a3+b3=1000-630
a3+b3=370

Answer 2

a + b = 10

a = b-10 .........1 )

Now,

a² + b² = 58

(b-10)² + b² = 58

b² + 100 - 20b + b² = 58

2b² - 20b + 42 = 0

2b² - 14b -6b +42=0

2b(b-7) -6(b-7)

(b - 7) (2b - 6)

So,

b = 7 ; 3

Now,

a³ + b³ :-

1 ) 7³+3³

= 370 ...............★ ANS ★


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