analytically? That is, without using numerical methods lớn attain an approximate solution.

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By the rational root test, this equation doesn"t have any rational roots.

So you have to lớn use the general method for solving cubic equations.

Using Cardano"s method (and a cas to vị the calculations):

\$\$x^3+x^2-4=0 ag1\$\$

and substitute \$\$x=t-1over3 ag2\$\$to get\$\$t^3-tover3-106over27=0 ag3\$\$

It is always possible khổng lồ find a substitution \$x=t+a\$ such that the coefficient of \$x^2\$ vanishes.

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Now we mix \$\$t=v+u ag4\$\$and get

\$\$v^3+left(v+u ight),left(9,u,v-1 ight)over3+u^3-106 over27=0 ag5\$\$

We impose the additional condition that \$\$u,v=1over9 ag6\$\$

and get

\$\$v^3+u^3=106over27 ag7\$\$

and

\$\$u^3,v^3=1over729 ag8\$\$

So \$u^3\$ & \$v^3\$ are the solution of the quadratic equation

\$\$z^2-106,zover27+1over729=0 ag9\$\$

This equation has the solutions

\$\$left -6,sqrt78-53over27 , 6,sqrt78+53 over27 ight ag10\$\$

\$u\$ & \$v\$ are interchangeable, we set

\$\$u^3=-6,sqrt78-53over27 ag11\$\$\$\$v^3=6,sqrt78+53 over27 ag12\$\$

Both values have a real cubic root. We can set

\$\$u=sqrt<3>-6,sqrt78-53over27 ag13\$\$\$\$v=sqrt<3>6,sqrt78+53 over27 ag14\$\$

because these solution pair satisfies \$(1)\$.

From this we get \$\$x=sqrt<3>-6,sqrt78-53over27 + sqrt<3>6,sqrt78+53 over27 +1/3 ag15\$\$

\$1\$ has three roots in \$usogorsk.combbC\$:\$\$left 1 , sqrt3,iover2-1over2 , -sqrt3,iover2-1over2 ight ag16\$\$So the two other soltuions we construct from

\$\$u=(sqrt<3>-6,sqrt78-53over27) (sqrt3,iover2-1over2) ag17\$\$\$\$v=(sqrt<3>6,sqrt78+53 over27) (-sqrt3,iover2-1over2) ag18\$\$

and

\$\$u=(sqrt<3>-6,sqrt78-53over27) (-sqrt3,iover2-1over2) ag19\$\$\$\$v=(sqrt<3>6,sqrt78+53 over27) (sqrt3,iover2-1over2) ag20\$\$