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avatar+11847 

Use suitable identity to get each of the following products!

 

vi. (a^2 + b^2) (-a^2 + b^2)

 

how do i solve it!no identity is matching with this!

rosala  Dec 28, 2014

Best Answer 

 #1
avatar+92221 
+15

(a^2 + b^2) (-a^2 + b^2)

 

$$(a^2 + b^2) (-a^2 + b^2)\\\\
=(b^2 + a^2) (b^2 -a^2)\\\\$$

 

Now you can see that it is  the difference of 2 squares. :)

so the answer is 

 

$$\\=(b^2)^2-(a^2)^2\\\\
=b^4-a^4$$

 

Does that all makes sense Rosala ?     

Melody  Dec 28, 2014
Sort: 

5+0 Answers

 #1
avatar+92221 
+15
Best Answer

(a^2 + b^2) (-a^2 + b^2)

 

$$(a^2 + b^2) (-a^2 + b^2)\\\\
=(b^2 + a^2) (b^2 -a^2)\\\\$$

 

Now you can see that it is  the difference of 2 squares. :)

so the answer is 

 

$$\\=(b^2)^2-(a^2)^2\\\\
=b^4-a^4$$

 

Does that all makes sense Rosala ?     

Melody  Dec 28, 2014
 #2
avatar+11847 
+10

umm no...not to me right now!

 

how did you  just swap everything ......i dont get it!

 

{i am talking about this step}(b^2 + a^2) (b^2 -a^2)

 

its like sorting out a ball  of wool!

 

rosala  Dec 28, 2014
 #3
avatar+92221 
+10

Okay rosala let's look at this.

a+b=b+a              agreed?     (1)

a-b=-b+a              agreed?     (2)

-b+a=a-b              agreed? (It is the same as the one above only in reverse)    (3)

so

$${{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{a}}}^{{\mathtt{2}}}$$            agreed?  (4)

and

$${\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}{\mathtt{\,\small\textbf+\,}}{{\mathtt{b}}}^{{\mathtt{2}}} = {{\mathtt{b}}}^{{\mathtt{2}}}{\mathtt{\,-\,}}{{\mathtt{a}}}^{{\mathtt{2}}}$$         agreed ?  (5)

 

so

 

$$(a^2 + b^2) (-a^2 + b^2)=(b^2 + a^2) (b^2 -a^2)\\\\$$       agreed  (6)

 

I have numbered all the lines so that you can tell me which ones don't make sense to you.  :)

Melody  Dec 28, 2014
 #4
avatar+11847 
+5

All of them make sense to me Melody!Than you very much!

 

rosala  Dec 28, 2014
 #5
avatar+92221 
+10

Good - I am glad i could help :)

Melody  Dec 28, 2014

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