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Question: How do you solve \(3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}\) ?...

How do you solve 3x3−2[x−(3−2x)]=3(x−2)23{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}} ?

Explanation

Solution

In this question, we have to find the value of x. To solve this problem, we will use algebraic identity, the distributive property, and the factoring method. We first use the algebraic identity (a−b)2=a2−2ab+b2{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} in the equation, then apply the distributive property a(b−c)=ab−aca(b-c)=ab-ac on the right-hand side of the equation. On further simplification, we will again apply the distributive property a(b−c)=ab−aca(b-c)=ab-ac on the left-hand side of the equation and hence get two separate equations. Thus, we will solve those equations separately, to get the required result for the problem.

Complete step-by-step solution:
According to the question, we have to find the value of x.
The equation given to us is 3x3−2[x−(3−2x)]=3(x−2)23{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}} --------- (1)
We will first apply the algebraic identity (a−b)2=a2−2ab+b2{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} on the right-hand side of the above equation, we get
3x3−2[x−(3−2x)]=3(x2−2.(x).(2)+22)3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-2.(x).(2)+{{2}^{2}} \right)
3x3−2[x−(3−2x)]=3(x2−4x+4)3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-4x+4 \right)
Now, we will apply the distributive property a(b−c)=ab−aca(b-c)=ab-ac on the right-hand side of the above equation, we get
3x3−2[x−(3−2x)]=3x2−3(4x)+3(4)3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{x}^{2}}-3(4x)+3(4)
Now, we will solve the brackets of the above equation, we get
3x3−2[x−3+2x]=3x2−12x+123{{x}^{3}}-2\left[ x-3+2x \right]=3{{x}^{2}}-12x+12
Now, we will apply the distributive property a(b−c)=ab−aca(b-c)=ab-ac on the left-hand side of the above equation, we get
3x3−2x−2(−3)−2(2x)=3x2−12x+123{{x}^{3}}-2x-2(-3)-2(2x)=3{{x}^{2}}-12x+12
On further solving, we get
3x3−2x+6−4x=3x2−12x+123{{x}^{3}}-2x+6-4x=3{{x}^{2}}-12x+12
Now, we will subtract 3x23{{x}^{2}} on both sides of the equation, we get
3x3−2x+6−4x−3x2=3x2−12x+12−3x23{{x}^{3}}-2x+6-4x-3{{x}^{2}}=3{{x}^{2}}-12x+12-3{{x}^{2}}
As we know, the same terms with opposite signs cancel out each other, therefore we get
3x3−2x+6−4x−3x2=−12x+123{{x}^{3}}-2x+6-4x-3{{x}^{2}}=-12x+12
On further simplification, we get
3x3−6x+6−3x2=−12x+123{{x}^{3}}-6x+6-3{{x}^{2}}=-12x+12
Now, we will subtract 1212 on both sides of the equation, we get
3x3−6x+6−3x2−12=−12x+12−123{{x}^{3}}-6x+6-3{{x}^{2}}-12=-12x+12-12
As we know, the same terms with opposite signs cancel out each other, therefore we get
3x3−6x−3x2−6=−12x3{{x}^{3}}-6x-3{{x}^{2}}-6=-12x
Now, we will add 12x12x on both sides of the equation, we get
3x3−6x−3x2−6+12x=−12x+12x3{{x}^{3}}-6x-3{{x}^{2}}-6+12x=-12x+12x
As we know, the same terms with opposite signs cancel out each other, therefore we get
3x3+6x−3x2−6=03{{x}^{3}}+6x-3{{x}^{2}}-6=0
Therefore, we get
3x3−3x2+6x−6=03{{x}^{3}}-3{{x}^{2}}+6x-6=0 ------ (2)
Now, we see that the above equation is in the form of a cubic equation, thus we will first separate equation (2) into two polynomials, we get
(3x3−3x2)+(6x−6)=0(3{{x}^{3}}-3{{x}^{2}})+(6x-6)=0
Now, we will take 3x23{{x}^{2}} common from the first polynomial and 6 from the next polynomial, we get
3x2(x−1)+6(x−1)=03{{x}^{2}}(x-1)+6(x-1)=0
So, we will take common (x-1) from the above equation, we get
(x−1)(3x2+6)=0(x-1)(3{{x}^{2}}+6)=0
Therefore, we will get two different equations from the above equation, we get
(x−1)=0(x-1)=0 -------- (3)
(3x2+6)=0(3{{x}^{2}}+6)=0 ------- (4)
Now, we will solve equation (3), which is
(x−1)=0(x-1)=0
Now, add 1 on both sides of the above equation, we get
x−1+1=0+1x-1+1=0+1
As we know, the same terms cancel out each other, therefore we get
x=1x=1
Now, we will solve equation (4), which is
(3x2+6)=0(3{{x}^{2}}+6)=0
Now, subtract 6 on both sides of the above equation, we get
3x2+6−6=0−63{{x}^{2}}+6-6=0-6
As we know, the same terms cancel out each other, therefore we get
3x2=−63{{x}^{2}}=-6
Now, we will divide 3 on both sides of the equation, we get
3x32=−63{{\dfrac{3x}{3}}^{2}}=\dfrac{-6}{3}
Therefore, we get
x2=−2{{x}^{2}}=-2
On further solving, we get
x=±i2x=\pm i\sqrt{2}
Therefore, for the equation 3x3−2[x−(3−2x)]=3(x−2)23{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}} ,the value of x is 1,+i2,−i21,+i\sqrt{2},-i\sqrt{2} .

Note: While solving this problem, keep in mind the steps you are using to avoid confusion and mathematical errors. One of the alternative methods to solve this problem of the cubic equation is by using the hit and trial method, after that, we will use the long division method to get a quadratic equation, and in the end, solve the quadratic equation using splitting the middle term method, to get the solution.