+0  
 
0
507
2
avatar

If a + b + c =11 and ab + bc + ac = 25,then find the value of a^3 + b^3 + c^3 – 3abc

 Jun 15, 2020
 #1
avatar+9466 
0

Notice that \(a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc +ca))\)

 

The only missing info we need is the value of \(a^2 + b^2 + c^2\).

 

Now, because \((a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\), plugging in values gives \(a^2 + b^2 + c^2 = 11^2 - 2(25) = 71\)

 

You can plug in the values on your own.

 Jun 15, 2020
 #2
avatar+1262 
+1

so like what maxwong said we have a^2+b^2+c^2=71 and a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - (ab + bc +ca))  plug that in to get 11*(71-25) and that equals 506.

 

 

 

 

 

Credts: Maxwong for hints

 Jun 15, 2020

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