Question

What is the value of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\] if \[a+b+c=12\] and \[ab+bc+ac=47\]?
(A) \[13\]
(B) \[36\]
(C) \[39\]
(D) \[32\]

Answer 1

Hint: To solve these kinds of questions we should know every formula related to algebra. This question is based on the concept of algebra and you should have good knowledge of algebra. The formula of algebra is used in solving various equations and inequalities in mathematics.

Complete step-by-step solution:
We use algebra for adding, subtracting, multiplying, and dividing various equations in mathematics. If we have to find the value of the missing equation and the value of some equation is given then by applying various formulas of algebra we can find out the missing equation. Alphabetical letters are used in algebra to find the missing number.
In algebra, like terms are those terms that have the same variable and exponent and unlike terms are those terms that have different exponents and variables.
In the above question, the value of the following equation is given.
\[\begin{align}
  & {{(12)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(47) \\
 & \Rightarrow 144={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+94 \\
 & 50={{a}^{2}}+{{b}^{2}}+{{c}^{2}} \\
\end{align}\]
\[a+b+c=12\]……….eq(1)
\[ab+bc+ac=47\]…….eq(2)
We have to find the value of \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
We know that,
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)\]…….eq(3)
Now first we will prove eq(3) and then we will proceed further with our question.
We will take RHS of our eq(3)
We will multiply each term of the first polynomial with the second polynomial.
\[(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)=a({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)+b({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)+c({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)\]
\[\Rightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)={{a}^{3}}+a{{b}^{2}}+a{{c}^{2}}-{{a}^{2}}b-abc-{{a}^{2}}c+{{a}^{2}}b+{{b}^{3}}+b{{c}^{2}}-a{{b}^{2}}-{{b}^{2}}c-abc+{{a}^{2}}c+{{b}^{2}}c+{{c}^{3}}-abc-b{{c}^{2}}-{{c}^{2}}a\]
After multiplying following results will be obtained
\[\Rightarrow (a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+{{a}^{2}}b-{{a}^{2}}b+a{{c}^{2}}-a{{c}^{2}}+a{{b}^{2}}-a{{b}^{2}}+b{{c}^{2}}-b{{c}^{2}}+{{a}^{2}}c-{{a}^{2}}c+{{b}^{2}}c-{{b}^{2}}c-abc-abc-abc\]
Equal and opposite terms will cancel out each other and we will get the following results as shown below.
\[(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-ab-bc-ac)={{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc\]
By this, the following identity will be proved.
We also know that,
\[{{(a+b+c)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(ab+bc+ac)\]…….eq(4)
On putting values of eq(1) and eq(2) in eq(4), the following results will be obtained as shown below.
\[{{(12)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2(47)\]
\[\Rightarrow 144={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+94\]
Now we will bring \[94\] on the left-hand side and we will subtract it from \[144\]and the following results will be obtained.
\[50={{a}^{2}}+{{b}^{2}}+{{c}^{2}}\]…..eq(5)
So the value if \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}\] comes out to be \[50\]
Now we will put the value of eq(5), eq(1), and eq(2) in eq(3), and the following results will be obtained.
\[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(a+b+c)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}-(ab+bc+ac))\]
\[\begin{align}
  & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=(12)(50-47) \\
 & \Rightarrow {{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=36 \\
\end{align}\]
So the correct option will be option(B) \[36\].

Note: When we add two numbers and both of them are negative then they both will be added but the sign of the resulting number will remain negative. When two numbers with opposite signs are multiplied with each other then the resulting number will be a number with a negative sign.