LOLS me again (1 Viewer)

[super.muppy]

someone please do (t-2)^3+(t+2)^3
Factor as fully as possible

I really did try

TY in advance

 

[speedofsound]

The answer is 4.

 

[supercalamari]

I'm SO glad I'm not doing maths this year, 160/180 kids in my year are already freaked out by their work... I know this doesn't answer your question, hope the smarter kids here can help you out.

 

[icecoffee]

2t(t^2 + 12)

i think...

 

[duckcowhybrid]

Let t-2=x
Let t+2=y

Therefore, you have

x^3 + y^3
= (x + y)(x^2 - xy + y^2) Now sub in original values

= (t-2 + t+ 2)(t^2 - 4t + 4 - (t^2 -4) + t^2 + 4t + 4)

= 2t(t^2 + 12)

And thats it. Not hard really, except typing it out on a computer. Same asnwer as above, but with working.

 

[icecoffee]

lol i was too lazy to type it out. ><

 

[bored of sc]

someone please do (t-2)^3+(t+2)^3
Factor as fully as possible

I really did try

TY in advance

It's an addition of two cubes.

I.e. a³+b³ = (a+b)(a²-ab+b²) but a = t-2 and b = t+2.

So (t-2)³+(t+2)³
= [(t-2)+(t+2)][(t-2)²-(t-2)(t+2)+(t+2)²]
= above answers

Just realised...

 

[Miss Sunshine]

bored of sc is right

it is the sum of two cubes

you need to know the rule

 

[lolokay]

you could also just expand them both

 

[TBK11]

i got bored lol
fully maxed out worked [=