Question

ifa²+b²=58and a+b =10 so find a³+b³+c³​

Answer 1

Answer:

Explanation:

(a+b)² = a²+b²+2ab.

(10)²=58+2ab

ab=(100-58)/2

ab=42/2=21

now,

(a+b)³=a³+b³+3ab(a+b)

10³=a³+b³+(3x21x10)

a³+b³=1000-630

therefore,a³+b³=370.