Question

if a+b+c =5 and ab+bc+ca = 10 then prove that a^3+b^3+c^3 = -25
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Answer 1

Step-by-step explanation:

Given,

a+b+c = 5

ab+bc+ca = 10

To proof that a³+b³+c³-3abc = -25

Proof: Using idenetity;

(a+b+c)² = a²+b²+c²+2ab+2bc+2ac

=> (a+b+c)² = a²+b²+c²+2(ab+bc+ac) [By taking 2 as common]

Subtituting the values given in the question,

=> (5)² = a²+b²+c²+2(10)

=> 25 = a²+b²+c²+20

=> 25-20 = a²+b²+c²

=> 5 = a²+b²+c²

So, we get a²+b²+c² = 5

Now, take the identity,

a³+b³+c³-3abc = (a+b+c)(a²+b²c²-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 = (5)(-5)

=> a³+b³+c³-3abc = -25

Hence, proved.