My 4th Problem

$27x^{3} + 21x + 8 = 0$ $27x^{3} + 1 + 21x + 7 = 0$ $(3x)^{3} + (1)^{3} + 21x + 7 = 0$ $(3x + 1) (9x^{2} - 3x + 1) + 7 (3x + 1) = 0$ $(3x + 1) (9x^{2} - 3x + 8) = 0$ Real root exist for $3x + 1 = 0$ $x = -0.33333$

Here is what I did,Upvote if you are satisfied!

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$27x^3+21x+8=0$

$(3x)^3+(2)^8+21x=0$

This can be written as

$(3x)^3+(2)^3+54x^2+15x+21x = 54x^2+15x$

$(3x+2)^3 = 54x^2+15x$ --------------Equation $1$

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Let $(3x+2) = u$

$9x^2+4+12x = u^2 \rightarrow\ 54x^2+72x+24 = 6u^2$

$\rightarrow$ $54x^2+15x = 6u^2-57x-24$

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$57x+38 = 19u$

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From equation $1$ $(3x+2)^3 = 54x^2+15x$

$\rightarrow$ $u^3 = 6u^2-7x-24$

$\rightarrow$ $u^3=6u^2-19u+14$

$\rightarrow$ $u^3-6u^2+19-14 = 0$

$\huge One \ root \ is \ 1$

$(u-1)(u^2-5u+14)$ = $u^3-6u^2+19u-14$

So roots come out to be

$u=1,\dfrac{5\pm\sqrt{-31}}{2}$

solving for $x$ from $u$

$\rightarrow$ $3x+2=1 \implies x= -1/3$

$\rightarrow$ $3x+2 = \dfrac{5\pm\sqrt{-31}}{2}$

$\implies$ $x=\dfrac{1\pm\sqrt{-31}}{6}$

so real roots are $\boxed{\dfrac{-1}{3}=-0.333.....}$

As per remainder theorem, $X=\dfrac{1,2,4,8}{1,3,9}.$ . Signs do not change so there is no +tive root. So trying f(-1,-1/3,-1/9....) we get $f(-\dfrac{1}{3}) =0.\\ \dfrac{f(X)}{3X+1} =9X^2-3X+8...~~~~But.9-4*8*7<0..\therefore~ there~ are~ no~\\ more~ real~ roots. ~~~~~~Answer ~~\boxed{\Large -0.333}$

${ 27x }^{ 3 }+21x+8=0\\ (3x+1)({ 9x }^{ 2 }-3x+8)=0\\ 3x+1=0$

or

${ 9x }^{ 2 }-3x+8 = 0$

But since ${ 9x }^{ 2 }-3x+8 = 0$ doesn't have a real root,

$x=-\frac { 1 }{ 3 } = \sim - 0.333$