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visits member for 2 years, 2 months
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I was born to answer lhfs.


May
18
comment Prove every odd integer is the difference of two squares
Would the downvoter care to explain?
May
18
awarded  Constituent
May
13
comment Is it an abuse of notation to omit the leading zero in a decimal less than 1?
@Gustavo It's Marvis. Haha
May
10
comment
Jayesh has a yes from my side!
May
7
awarded  Caucus
Apr
21
comment An infinite fraction
Perfect observation. How'd you think of that?
Apr
19
revised With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$
deleted 14 characters in body
Apr
19
revised Is the following statement true? Why?
added the important part
Apr
19
answered Is the following statement true? Why?
Apr
19
awarded  Cleanup
Apr
19
comment With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$
@jack $[(3 - \sqrt{2})^2]^2 = [11 - 6\sqrt2]^2$ because $11 - 6\sqrt2 = (3 - \sqrt{2})^2$
Apr
19
revised With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$
rolled back to a previous revision
Apr
19
answered With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$
Apr
19
comment With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$
forgetting $\binom{4}{k}$?
Apr
19
answered Derivative of $\sin(e^{-x})$
Apr
14
accepted Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$
Apr
14
comment Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$
Nice alternative. Thanks.
Apr
14
comment Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$
@Easy wow, that is epic! Gotcha!
Apr
14
comment Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$
$= -(b+a)(3c^2 - 3bc + b^2 + ac - ba + 2ac + a^2)$
Apr
14
comment Factoring a long expression in the form $(a+b)^3 + (c - b)^3 - (c+b)^3$
@Jay Ah, I didn't notice that.