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Answer 1 · 2018-01-31T14:09:12

#(a-4)^3 = (a-4)(a-4)(a-4)#

#(a-4)^3 = (a-4)(a-4)(a-4)#

# = (a^2-8a+16)(a-4)#

# = a^3-12a^2+48a-64#

So, #(a-4)^3 = a^3-12a^2+48a-69#

Numerical examples

For #a = 5#,

#(a-4)^3 = (5-4)^3 = 1^3 = 1#

But #a^3-64 = 5^3-64 = 125 - 64 = 61#

For #a = 1#

#(a-4)^3 = (1-4)^3 = (-3)^3 = -27#

But #a^3-64 = (-1)^3-64 = -1 - 64 = -65#

One way to think about why #(a-4)^3# is NOT #a^3-64# is to think about order os operations. We have to do the subtraction before the exponent.