Ace school

with brainly

  • Get help from millions of students
  • Learn from experts with step-by-step explanations
  • Level-up by helping others

A community for students.

why doesnt x^3-8=(x-2)^3??? please reply!

Mathematics
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Join Brainly to access

this expert answer

SEE EXPERT ANSWER

To see the expert answer you'll need to create a free account at Brainly

becasue they are 2 different operations
exponents are(nt) distributive across addition
In one you are cubing the x and subtracting the 8. In the other you are subtracting the 2 then cubing it. On a side not: I WANNA BE A MOD AMISTRE!!!

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

because (x-2)^3 is like (x-2)(x-2)(x-2)
so, it doesn't equal the same as x^3-8
AMISTRE!!! D: My lip is poking out :(
Multiply out \[(x-2)^3\] and you will see that the answer is not \[x^3-8\]
In general: \[(a+b)^n \neq a^n+b^n\]
but how would u know it is (x-2)(x^2+2x+4)
We agree that 2^3=2*2*2? So: (x-2)^3=(x-2)(x-2)(x-2) So you can get: (x-2)(x^2+2x+4)
ohhh i understand thanks so muh guys!

Not the answer you are looking for?

Search for more explanations.

Ask your own question