Question

without actually calculating the cubes, find the value of -(0.4)cube - (0.2)cube + (0.6)cube

Answer 1

identity a3+b3+c3-3abc=(a+b+c)(a2+b2+c2-ab-bc-ca)
-0.4-0.2+0.6=0.
(-0.4)3+(-0.2)3+(0.6)3=0.144.

the answer is 0.144

hope it helps.
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Answer 2

- (0.4)³ - (0.2)³ + (0.6)³ = 0.144

Given :

- (0.4)³ - (0.2)³ + (0.6)³

To find :

The value of the expression

Formula :

If a + b + c = 0 then a³ + b³ + c³ = 3abc

Solution :

Step 1 of 2 :

Find the value of (- 0.4) + ( - 0.2) + 0.6

Let a = - 0.4 , b = - 0.2 , c = 0.6

Then we have

a + b + c

= - 0.4 - 0.2 + 0.6

= 0

Step 2 of 2 :

Find the value of the expression

We know that if a + b + c = 0 then a³ + b³ + c³ = 3abc

Thus we get

- (0.4)³ - (0.2)³ + (0.6)³

= ( - 0.4)³ ( - 0.2)³ + (0.6)³

= a³ + b³ + c³

= 3abc

= 3 × ( - 0.4 ) × ( - 0.2) × 0.6

= 0.144

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