Question

without actually calculating the cubes find 2(0.3)³+(0.4)³+(0.5)³+(-0.7)³+(-0.8)³

Answer 1

2(0.3)³ + (0.4)³ + (0.5)³ + (-0.7)³ + (-0.8)³
= (0.3)³ + (0.3)³ + (0.4)³ + (0.5)³ + (-0.7)³ + (-0.8)³
= (0.3)³ + (0.4)³ + (-0.7)³ + (0.3)³ +  (0.5)³ + (-0.8)³
= [(0.3)³ + (0.4)³ + (-0.7)³] + [(0.3)³ +  (0.5)³ + (-0.8)³]  ------------------(1)

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

In [(0.3)³ + (0.4)³ + (-0.7)³]; 
0.3 + 0.4 + (-0.7) = 0.7 -0.7 = 0
So (0.3)³ + (0.4)³ + (-0.7)³
= 3×0.3×0.4×(-0.7) 
⇒ [(0.3)³ + (0.4)³ + (-0.7)³] = -0.252       ------------------------------------(2)

Similarly In [(0.3)³ +  (0.5)³ + (-0.8)³]; 
0.3 + 0.5 + (-0.8) = 0.8 -0.8 = 0
So (0.3)³ + (0.4)³ + (-0.7)³ 
= 3×0.3×0.5×(-0.8) 
⇒ [(0.3)³ +  (0.5)³ + (-0.8)³] = -0.360       ------------------------------------(3)

So putting equation (2) and (3) in equation (1)
[(0.3)³ + (0.4)³ + (-0.7)³] + [(0.3)³ +  (0.5)³ + (-0.8)³] 
= (-0.252) + (-0.360)
= -0.612

So 2(0.3)³ + (0.4)³ + (0.5)³ + (-0.7)³ + (-0.8)³ = -0.612

Answer 2

 we use the Algebra formula : a³ + b³ + c³ = 3 a b c    if a+b+c =0
  we observe that  0.3 + 0.4 + (-0.7) = 0
                 and ,  0.3 + 0.5 + (-0.8) = 0

LHS = 0.3³ + 0.4³ + (-0.7)³ + 0.3³ + 0.5³ + (-0.8)³
       = 3 * 0.3 * 0.4 * (0.7)  +  3 * 0.3 * 0.5 * (-0.8)
       =  - 0.9 * (0.28 + 0.40)
       = - 0.612