## kaylala 2 years ago 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.

1. kaylala

2. kaylala

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

3. chmvijay

can you expand it by multiplication

4. shkrina

r there powers

5. kaylala

i dont know how? @chmvijay

6. shkrina

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

7. shkrina

IS this ur question \[((x^{2}+3y)+y ^{2})((x ^{2}+3y)-y ^{2}\])

8. shkrina

hellooo.........

9. kaylala

yes, it is @shkrina

10. shkrina

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

11. kaylala

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

12. whpalmer4

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

13. kaylala

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

14. kaylala

???????

15. shkrina

@kaylala did u finish this sum....

16. kaylala

not sure @shkrina

17. shkrina

well .|dw:1372577408636:dw|

18. shkrina

u will get \[x ^{4}+3x ^{2}y-x ^{2}y ^{2}+3yx ^{2}+9y ^{2}-3y3+x ^{2}y ^{2}+3y ^{3}-y ^{4}\]

19. shkrina

simplify u will get the ans ///

20. kaylala

@shkrina but this is my answer: x4+6x2−y4+9y2

21. shkrina

ss rt ans ....the above eq is the eq u get after multiplying ....u can cancle 3y^3and x^2y^2/.....and add ..3x^2y ... u will get the same ans ....hope clear .....

22. shkrina

@kaylala

23. whpalmer4

@shkrina you disregarded the instructions for this problem. Whether or not there's a different way to solve a problem, if the instructor asks for it to be done a specific way, that's the way it should be done for the purposes of learning the material. @kaylala as I said \[(a+b)(a-b) = a^2-b^2\] with \(a=(x^2+3y),b=y^2\) Write out \[(a+b)(a-b) = a^2-b^2\]That is the "special product" Now replace \(a\) on the right hand side with \((x^2+3y)\): \[a^2-b^2 = (x^2+3y)^2 -b^2\]Now replace \(b\) on the right hand side with \(y^2\) \[a^2-b^2=(x^2+3y)^2-(y^2)^2\] That was the "chain rule" of special product Now we have to expand everything to get rid of the grouping symbols. Let's do the easy one first: \[(x^2+3y)^2-(y^2)^2 = (x^2+3y)^2-y^4\]because \[(u^n)^m = u^{n*m}\]Now we expand the first product term by repeatedly applying the distributive property \[(x^2+3y)^2-y^4 = (x^2+3y)(x^2+3y)-y^4 = x^2(x^2+3y)+3y(x^2+3y) - y^4\]\[=x^2*x^2+x^2*3y + 3y*x^2+3y*3y - y^4 \]\[=x^4+3x^2y+3x^2y+9y^2-y^4\]Now collect like terms \[=x^4+6x^2y+9y^2-y^4\] So your answer was very close, but you somehow lost the \(y\) portion of the \(6x^2y\) term. You should revisit your work and try to figure out why you made that mistake.

24. shkrina

@kaylala ,, @whpalmer4 well sorry ....i did not pay much attention to the variable ...... this time i will look through........ once again sorry ,,,,,, @kaylala in u r ans 6x^2y ,,,,not 6x^2....u missed y...... hope its clear .........