Please Derive (4) from (3) please show step by step and show work How to Find the Exact Solution of a General Cubic EquationIn this chapter, we are going to find the exact solution of a general cubic equationax³ + bx² + cx+d=0(1)To find the roots of Equation (1), we first get rid of the quadratic term (x²) by making thesubstitutionx=y=to obtainb3awherebd{y− b )' + b[{y_b )² +c{y − 3 ) + d.+d=0Expanding Equation (3) and simplifying, we obtain the following equation- 3 day+[d+26³ be27a² 3a=0ay³ + C-26³ bcy² + ² [c-²y+² (d+ 2² -0=3ay³ + ey+f=0Equation (4) is called the depressed cubic since the quadratic term is absent. Having theequation in this form makes it easier to solve for the roots of the cubic equation (Click here toknow the history behind solving cubic equations exactly).First, convert the depressed cubic Equation (4) into the form26³ bc27a 3aNow, reduce the above equation using Vieta's substitutiony=z+=For the time being, the constant s is undefined. Substituting into the depressed cubicEquation (5), we get( ² + ² ) ² + √² + ² } + s =f=== (₁ +where we assumed.W=z³y=z+S=x=y(2)bЗа(3)=0(4)Expanding out and multiplying both sides by z³, we getzº +(3s +e)z¹ + f2³ + s(3s+e)2² +s³ = 0Now, let s= (s is no longer undefined) to simplify the equation into a tri-quadraticequation.25 + 12³ - 2/7 = 0(9)By making one more substitution, w=z³, we now have a general quadratic equation whichcan be solved using the quadratic formula.e³w² + fw-27Once you obtain the solution to this quadratic equation, back substitute using the previoussubstitutions to obtain the roots to the general cubic equation.wz→y →x(5)(6)(7)(8)(10)(11)(12)Note: You will get two roots for w as Equation (10) is a quadratic equation. UsingEquation (11) would then give you three roots for each of the two roots of w, hence givingyou six root values for z. But the six root values of z would give you six values of y(Equation (6)); but three values of y will be identical to the other three. So one gets onlythree values of y, and hence three values of x. (Equation (2))

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Equation (3):We have to prove equation (4) which is:Formula used:here, , and .Therefore,

+(x-1)+(x-1)+qy-2)+ Expanding Equation (3) and simplifying, we obtain the following equation +(c-512)+(a+; За За 2b be 27a2 за + d = 0 = =0 (3)