\[x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)+(4a)^2)+x^2\dotsm}}}\] \(Derivation:\)I made use of the following identity; \(x+a=\sqrt[3]{(x+a)^3}\\=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}\) In the 2nd step I wrote \(3ax^2\) as \(2ax^2+ax^2\).In the 3rd step I rearranged the terms.In the 4th step I factored out the \(a\) from the first 3 terms and I factored out the \(x^2\) from the last 2 terms.Manipulating \(x+2a=\sqrt[3]{x^3+3(2a)x^2+3(2a)^2x+(2a)^3}\) in the same way,I got \(x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}\).In the same way I got the expression for \(x+4a,x+8a,x+16a\dotsm\).Now I observed that each of these can be substituted in the previous expression.Substituting my way upto \(x+a\),I got the expression written at the beginning.Now,I want the people on Brilliant to verify this expression and to see whether it works.Please help me!!!

## Comments

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TopNewestPut a = 0

R.H.S becomes 0 but L.H.S is still 'x'

So I don't think it's correct – Krishna Sharma · 1 year, 11 months ago

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Of course, it could be that the statement is not true for \( a = 0 \), but is true for all other real numbers. – Calvin Lin Staff · 1 year, 11 months ago

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– Abdur Rehman Zahid · 10 months, 1 week ago

Just looking back at this note and thinking......That was a long time agoLog in to reply

– Aditya Sky · 6 months, 2 weeks ago

Putting \(a \,=\, 0\) yields another infinitely nested series of \(x\), which on simplification is equal to \(x\).Log in to reply

– Abdur Rehman Zahid · 6 months, 2 weeks ago

YupLog in to reply

– Abdur Rehman Zahid · 1 year, 11 months ago

Please explain how R.H.S becomes zeroLog in to reply

– Krishna Sharma · 1 year, 11 months ago

Every term on R.H.S is multiple of 'a' you can see itselfLog in to reply

– Abdur Rehman Zahid · 1 year, 11 months ago

If we rearrange the formula for \(x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\), you get \(x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}\) In this expression if you replace a by 2a, you get \(x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}\)Replacing this expression in the expression for x +a, you get \(x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a))}}\)In this way if we continue deriving the expression for x+4a,x+8a,x+16a etc and replacing them in the original formula for x+a we will get the original formula .So R.H.S does not become zero you just have to replace the value of x+2a instead of the next radical and then set a=0 and after simplifying you get \(x=\sqrt[3]{x^3}\)Log in to reply

Ah yes. Isn't it amazing how much can be achieved in a year? – Calvin Lin Staff · 10 months, 1 week ago

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– Abdur Rehman Zahid · 10 months ago

Yeah.I didn't even know a thing back then.Brilliant has helped me a lotLog in to reply

if a=0 then x also equal to 0 as x-x=0 – Muhammad Yousaf Khan · 1 year, 10 months ago

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