How do you solve $3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}$ ?
Answer
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Hint: 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 ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ in the equation, then apply the distributive property $a(b-c)=ab-ac$ on the right-hand side of the equation. On further simplification, we will again apply the distributive property $a(b-c)=ab-ac$ 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 $3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}$ --------- (1)
We will first apply the algebraic identity ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ on the right-hand side of the above equation, we get
$3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-2.(x).(2)+{{2}^{2}} \right)$
\[3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-4x+4 \right)\]
Now, we will apply the distributive property $a(b-c)=ab-ac$ on the right-hand side of the above equation, we get
\[3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{x}^{2}}-3(4x)+3(4)\]
Now, we will solve the brackets of the above equation, we get
\[3{{x}^{3}}-2\left[ x-3+2x \right]=3{{x}^{2}}-12x+12\]
Now, we will apply the distributive property $a(b-c)=ab-ac$ on the left-hand side of the above equation, we get
\[3{{x}^{3}}-2x-2(-3)-2(2x)=3{{x}^{2}}-12x+12\]
On further solving, we get
\[3{{x}^{3}}-2x+6-4x=3{{x}^{2}}-12x+12\]
Now, we will subtract $3{{x}^{2}}$ on both sides of the equation, we get
\[3{{x}^{3}}-2x+6-4x-3{{x}^{2}}=3{{x}^{2}}-12x+12-3{{x}^{2}}\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}-2x+6-4x-3{{x}^{2}}=-12x+12\]
On further simplification, we get
\[3{{x}^{3}}-6x+6-3{{x}^{2}}=-12x+12\]
Now, we will subtract $12$ on both sides of the equation, we get
\[3{{x}^{3}}-6x+6-3{{x}^{2}}-12=-12x+12-12\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}-6x-3{{x}^{2}}-6=-12x\]
Now, we will add $12x$ on both sides of the equation, we get
\[3{{x}^{3}}-6x-3{{x}^{2}}-6+12x=-12x+12x\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}+6x-3{{x}^{2}}-6=0\]
Therefore, we get
\[3{{x}^{3}}-3{{x}^{2}}+6x-6=0\] ------ (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
$(3{{x}^{3}}-3{{x}^{2}})+(6x-6)=0$
Now, we will take $3{{x}^{2}}$ common from the first polynomial and 6 from the next polynomial, we get
$3{{x}^{2}}(x-1)+6(x-1)=0$
So, we will take common (x-1) from the above equation, we get
$(x-1)(3{{x}^{2}}+6)=0$
Therefore, we will get two different equations from the above equation, we get
$(x-1)=0$ -------- (3)
$(3{{x}^{2}}+6)=0$ ------- (4)
Now, we will solve equation (3), which is
$(x-1)=0$
Now, add 1 on both sides of the above equation, we get
$x-1+1=0+1$
As we know, the same terms cancel out each other, therefore we get
$x=1$
Now, we will solve equation (4), which is
$(3{{x}^{2}}+6)=0$
Now, subtract 6 on both sides of the above equation, we get
$3{{x}^{2}}+6-6=0-6$
As we know, the same terms cancel out each other, therefore we get
$3{{x}^{2}}=-6$
Now, we will divide 3 on both sides of the equation, we get
${{\dfrac{3x}{3}}^{2}}=\dfrac{-6}{3}$
Therefore, we get
${{x}^{2}}=-2$
On further solving, we get
$x=\pm i\sqrt{2}$
Therefore, for the equation $3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}$ ,the value of x is $1,+i\sqrt{2},-i\sqrt{2}$ .
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.
Complete step-by-step solution:
According to the question, we have to find the value of x.
The equation given to us is $3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}$ --------- (1)
We will first apply the algebraic identity ${{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ on the right-hand side of the above equation, we get
$3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-2.(x).(2)+{{2}^{2}} \right)$
\[3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3\left( {{x}^{2}}-4x+4 \right)\]
Now, we will apply the distributive property $a(b-c)=ab-ac$ on the right-hand side of the above equation, we get
\[3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{x}^{2}}-3(4x)+3(4)\]
Now, we will solve the brackets of the above equation, we get
\[3{{x}^{3}}-2\left[ x-3+2x \right]=3{{x}^{2}}-12x+12\]
Now, we will apply the distributive property $a(b-c)=ab-ac$ on the left-hand side of the above equation, we get
\[3{{x}^{3}}-2x-2(-3)-2(2x)=3{{x}^{2}}-12x+12\]
On further solving, we get
\[3{{x}^{3}}-2x+6-4x=3{{x}^{2}}-12x+12\]
Now, we will subtract $3{{x}^{2}}$ on both sides of the equation, we get
\[3{{x}^{3}}-2x+6-4x-3{{x}^{2}}=3{{x}^{2}}-12x+12-3{{x}^{2}}\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}-2x+6-4x-3{{x}^{2}}=-12x+12\]
On further simplification, we get
\[3{{x}^{3}}-6x+6-3{{x}^{2}}=-12x+12\]
Now, we will subtract $12$ on both sides of the equation, we get
\[3{{x}^{3}}-6x+6-3{{x}^{2}}-12=-12x+12-12\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}-6x-3{{x}^{2}}-6=-12x\]
Now, we will add $12x$ on both sides of the equation, we get
\[3{{x}^{3}}-6x-3{{x}^{2}}-6+12x=-12x+12x\]
As we know, the same terms with opposite signs cancel out each other, therefore we get
\[3{{x}^{3}}+6x-3{{x}^{2}}-6=0\]
Therefore, we get
\[3{{x}^{3}}-3{{x}^{2}}+6x-6=0\] ------ (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
$(3{{x}^{3}}-3{{x}^{2}})+(6x-6)=0$
Now, we will take $3{{x}^{2}}$ common from the first polynomial and 6 from the next polynomial, we get
$3{{x}^{2}}(x-1)+6(x-1)=0$
So, we will take common (x-1) from the above equation, we get
$(x-1)(3{{x}^{2}}+6)=0$
Therefore, we will get two different equations from the above equation, we get
$(x-1)=0$ -------- (3)
$(3{{x}^{2}}+6)=0$ ------- (4)
Now, we will solve equation (3), which is
$(x-1)=0$
Now, add 1 on both sides of the above equation, we get
$x-1+1=0+1$
As we know, the same terms cancel out each other, therefore we get
$x=1$
Now, we will solve equation (4), which is
$(3{{x}^{2}}+6)=0$
Now, subtract 6 on both sides of the above equation, we get
$3{{x}^{2}}+6-6=0-6$
As we know, the same terms cancel out each other, therefore we get
$3{{x}^{2}}=-6$
Now, we will divide 3 on both sides of the equation, we get
${{\dfrac{3x}{3}}^{2}}=\dfrac{-6}{3}$
Therefore, we get
${{x}^{2}}=-2$
On further solving, we get
$x=\pm i\sqrt{2}$
Therefore, for the equation $3{{x}^{3}}-2\left[ x-\left( 3-2x \right) \right]=3{{\left( x-2 \right)}^{2}}$ ,the value of x is $1,+i\sqrt{2},-i\sqrt{2}$ .
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.
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