Look for the special product of:

[( x2 + 3y) + y2] [(x2 + 3y) – y2]

NOTE:

~ See the clearer given in the comment area

~ Group some terms to find the applicable special product.
~Follow the chain rule of special product.
~ The Final answer should be free from all symbols of grouping.

Question

Look for the special product of:

[( x2 + 3y) + y2] [(x2 + 3y) – y2]



NOTE:

~ See the clearer given in the comment area


~ Group some terms to find the applicable special product. 
~Follow the chain rule of special product.
~ The Final answer should be free from all symbols of grouping.

kaylala · Answer

attachment

kaylala · Answer

See the document posted above for a CLEARER VIEW of the GIVEN question

chmvijay · Answer

can you expand it by multiplication

anonymous · Answer

r there powers

kaylala · Answer

i dont know how?

@chmvijay

anonymous · Answer

Well  first add the like terms .. in both groups ...

anonymous · Answer

IS this ur question $$((x^{2}+3y)+y ^{2})((x ^{2}+3y)-y ^{2}$$)

anonymous · Answer

hellooo.........

kaylala · Answer

yes, it is

@shkrina

anonymous · Answer

well if so multiply every term of group ((x2+3y)+y2) with all elements of  group ((x2+3y)+y2) .... done

kaylala · Answer

so, what's the product or the final answer?

whpalmer4 · Answer

My guess is that you are supposed to think of this as $$(a+b)(a-b) = a^2-b^2$$with $a = (x^2+3y) , b = y^2$

kaylala · Answer

ok @whpalmer4 

is the answer equal to:

$$x ^{4}+6x ^{2}-y ^{4}+9y ^{2}$$

kaylala · Answer

???????

anonymous · Answer

@kaylala  did u finish this sum....

kaylala · Answer

not sure @shkrina

anonymous · Answer

well  .|dw:1372577408636:dw|