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this is part of the set @the Radicals \[\] let there be a infinite Radical \[ \sqrt{17+2\sqrt{88+24\sqrt{17+2\sqrt{88+24\sqrt{\dots\dots }}}}}\] it can be written as \[\sqrt{3^2+2^2+2^2+2\sqrt{(3\times2)^2+(2\times 2)^2 +(3\times 2)^2 +2(3\times 2\times 2)\sqrt{\dots\dots }}}\] then we can say that \[ \sqrt{17+2\sqrt{88+24\sqrt{17+2\sqrt{88+24\sqrt{\dots\dots }}}}}=7\] how??? lets see\[\] let \[F(a,b,c)=a+b+c\] \[F(a,b,c)=\sqrt{a^2 +b^2+c^2 +2(ab+bc+ca)}\] \[F(a,b,c)=\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2(a^2bc+ab^2c+abc^2)}}\] \[F(a,b,c)=\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc(a+b+c)}}\] now we can expand $a+b+c$ like before and make it an infinite radical\[\] so we can say that\[\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{a^2 +b^2 +c^2 +2\sqrt{(ab)^2+(bc)^2+(ca)^2+2abc\sqrt{\dots\dots}}}}} =a+b+c\]

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Comments

$x+a=\sqrt[3]{(x+a)^3}\\x+a=\sqrt[3]{x^3+3ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{x^3+2ax^2+ax^2+3a^2x+a^3}\\x+a=\sqrt[3]{ax^2+3a^2x+a^3+x^3+2ax^2}\\x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2(x+2a)}$ In the same way get $x+2a=\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2(x+4a)}$ and $x+4a=\sqrt[3]{4a(x^2+3(4a)x+(4a)^2)+x^2(x+8a)}$ and the expressions for $x+8a,x+16a\;etc$ .Replacing all of these in the original expression we get: $\color{#D61F06}{x+a=\sqrt[3]{a(x^2+3ax+a^2)+x^2\sqrt[3]{2a(x^2+3(2a)x+(2a)^2)+x^2\sqrt[3]{4a(x^2+3(4a)x+(4a)^2)+x^2\sqrt[3]{\dotsm\;to\; \infty}}}}}$ $\LARGE{\color{#20A900}{V}\color{#3D99F6}{O}\color{#624F41}{I}\color{#69047E}{L}\color{#D61F06}{A}\color{#333333}{!!!}}$