Question

if a+b+c=5, ab+bc+ca=3 find the value of a³+b³+c³-3abc​

Answer 1

❍ GIVEN:

a+b+c = 5

ab+bc+ca = 3

➢ TO FIND: value of a³+b³+c³-3abc

✯ SOLUTION:

❍ We know that:

a³+b³+c³-3abc =(a+b+c)(a²+c²+c²-ab-bc-ca) ...(i)

then, first we want a²+b²+c²

so,

⟹ (a+b+c)²= a²+b²+c²+2(ab+bc+ca)

⟹ (5)² = a²+b²+c²+2(3)

⟹ 25 = a²+b²+c²+6

∴ a²+b²+c² = 19

now we want:

➢ (-ab-bc-ca)

⟹ given:

⟹ (ab+bc+ca) = 3

⟹ -3 = - (ab+bc+ca)

∴ (-ab-bc-ca) = -3

➢ Now, we will find a³+b³+c³-3abc

by using the indentity ....(i)

⟹ a³+b³+c³-3abc = (5) (19- (-3))

⟹ a³+b³+c³-3abc= 5× 22

∴ a³+b³+c³-3abc = 110