Question

please answer this question correctly quality answer needed​

Answer 1

Step-by-step explanation:

Given :-

a + b + c = 5

ab + bc + ca = 10

To find :-

Prove that a³ + b³ + c³ - 3abc = -25

Solution :-

Given that

a + b + c = 5 ---------------(1)

ab + bc + ca = 10 ---------(2)

On squaring equation (1) both sides then

=> (a+b+c)² = 5²

=> a²+b²+c²+2ab+2bc+2ca = 25

=> a²+b²+c²+2(ab+bc+ca) = 25

On Substituting the value of ab+bc+ca = 10 then

=> a²+b²+c²+2(10) = 25

=> a²+b²+c²+20 = 25

=> a²+b²+c² = 25-20

=> a²+b²+c² = 5 ----------------(3)

We know that

(a+b+c)(a²+b²+c²-ab-bc-ca) = a³+b³+c³-3abc

=> (a+b+c)((a²+b²+c²-(ab+bc+ca)) = a³+b³+c³-3abc

=> (5)(5-(10)) = a³+b³+c³-3abc

From (1),(2)&(3)

=> 5(5-10) = a³+b³+c³-3abc

=> 5(-5) = a³+b³+c³-3abc

=> -25 = a³+b³+c³-3abc

=> a³+b³+c³-3abc = -25

Hence , Proved .

Answer :-

If a + b + c = 5, ab + bc + ca = 10 then a³+b³+c³-3bc = -25

Used formulae:-

• (a+b+c)² = a²+b²+c²+2ab+2bc+2ca

• a³+b³+c³-3abc= (a+b+c)(a²+b²+c²-ab-bc-ca)