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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)
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