Question #a7e02

2 Answers
Nov 29, 2016

Answer:

Actually, Josiah is correct

Explanation:

If #x# was equal to zero then the two expressions will be equal

If #x=0#

1) #(-2x)^3=(-2*0)^3=0^3=0#

2) #-2x^3=-2*0^3=-2*0=0#

But if it was any number else then the expressions won't be equal

If #x=1#

1) #(-2*1)^3=(-2)^3=-8#

2) #-2*1^3=-2*1=-2#

This is because the first expression was raised to the power of #3#

#(-2x)^3=(-2x)(-2x)(-2x)=(-2)(-2)(-2)(x)(x)(x)=-8x^3#

And the second expression is

#-2x^3#

#-8x^3# is not the same as #-2x^3#, so they will produce different values, except for #x=0#

Nov 30, 2016

Answer:

They are the same if #x=0# otherwise not the same

Explanation:

#color(blue)("Consider: "(-2x)^3)#

The brackets group together the -2 and the #x#

So the index (power) is applied to everything inside the bracket giving:

#(-2x)xx(-2x)xx(-2x)#

This is the same as:

#(-2)^3xx(x)^3 = -8x^3#

So #" "color(blue)((-2x)^3=-8x^3)#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

#color(blue)("Consider: "-2x^3#

In this case the -2 and the #x# are not 'locked' together other than by the operation of multiply. So we have:

#(-2)xxx^3" "=" "-2x^3#

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Comparing the two")#

It is stated that #(-2x)^3=-2x^3#

#=> -8x^3=-2x^3#

#color(brown)("Consider the case "x!=0")#

Divide both sies by #x^3# giving

#-8x^3=-2x^3" "->" " -8=-2 color(red)(larr" False") #

#"So for "x!=0:" " -8x^3!=-2x^3#
..................................................................................................

#color(brown)("Consider the case "x=0")#

For #x=0:" "-8x^3=02x^3#

#-8(0)^3=-2(0)^2#

#0=0" "color(red)(larr" True")#

#"So for "x=0:" " -8x^3=-2x^3#