If \[{{\text{a}}^3}{\text{ }}{\text{ + }}{{\text{b}}^3}{\text{ + }}{{\text{c}}^3}{\text{ = 3abc}}\] and \[{\text{a }}{\text{ + b + c = 0}}\] show that \[\dfrac{{{{{\text{(b + c)}}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{({\text{c + a)}}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{({\text{a + b)}}}^2}}}{{3 {\text{ab}}}}{\text{ = 1}}\]
Answer
635.7k+ views
Hint: We had to only find the value of b + c in terms of a, c + a in terms of b and a + b in terms of c from the given equation a + b + c = 0. Then we will put those values in LHS of the given equation we had to prove.
Complete step-by-step answer:
Now, it is given that \[{\text{a }}{\text{ + b + c = 0}}\] .We have to prove that LHS = R. H. S. Now from the given condition we will find the values of b + c, c + a, a + b and put the term in the left – hand side term.
As, a + b + c = 0, so b + c = -a, c + a = -b, a + b = -c.
Putting these values in the left – hand side, we get
L. H. S = \[\dfrac{{{{{\text{(b + c)}}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{({\text{c + a)}}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{({\text{a + b)}}}^2}}}{{3{\text{ab}}}}\]
$ \Rightarrow $ L. H. S = \[\dfrac{{{{{\text{( - a)}}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{({\text{ - b)}}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{( - {\text{c)}}}^2}}}{{3{\text{ab}}}}\]
As, ${({\text{ - x)}}^2}{\text{ = }}{{\text{x}}^2}$. Therefore, the left - hand side is
$ \Rightarrow $ L. H. S = \[\dfrac{{{{\text{a}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{\text{b}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{\text{c}}^2}}}{{3{\text{ab}}}}\]
Now, taking 3abc as lcm in the above equation, we get
$ \Rightarrow $ L. H. S = \[\dfrac{{{{\text{a}}^3}{\text{ + }}{{\text{b}}^3}{\text{ + }}{{\text{c}}^3}}}{{3{\text{abc}}}}\]
As given in the question, \[{{\text{a}}^3}{\text{ }}{\text{ + }}{{\text{b}}^3}{\text{ + }}{{\text{c}}^3}{\text{ = 3abc}}\]. Applying this condition in the above equation
$ \Rightarrow $ L. H. S = \[\dfrac{{{\text{3abc}}}}{{3{\text{abc}}}}\] = 1 = R. H. S
Hence proved.
Note: To solve such types of questions, we have to follow a few steps. First, we will find the value of the variable which we will use to solve the problem. Then we will use the given condition in the question and put the value of the variable in the given condition to solve the given problem.
Complete step-by-step answer:
Now, it is given that \[{\text{a }}{\text{ + b + c = 0}}\] .We have to prove that LHS = R. H. S. Now from the given condition we will find the values of b + c, c + a, a + b and put the term in the left – hand side term.
As, a + b + c = 0, so b + c = -a, c + a = -b, a + b = -c.
Putting these values in the left – hand side, we get
L. H. S = \[\dfrac{{{{{\text{(b + c)}}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{({\text{c + a)}}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{({\text{a + b)}}}^2}}}{{3{\text{ab}}}}\]
$ \Rightarrow $ L. H. S = \[\dfrac{{{{{\text{( - a)}}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{({\text{ - b)}}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{( - {\text{c)}}}^2}}}{{3{\text{ab}}}}\]
As, ${({\text{ - x)}}^2}{\text{ = }}{{\text{x}}^2}$. Therefore, the left - hand side is
$ \Rightarrow $ L. H. S = \[\dfrac{{{{\text{a}}^2}}}{{3{\text{bc}}}}{\text{ + }}\dfrac{{{{\text{b}}^2}}}{{{\text{3ac}}}}{\text{ + }}\dfrac{{{{\text{c}}^2}}}{{3{\text{ab}}}}\]
Now, taking 3abc as lcm in the above equation, we get
$ \Rightarrow $ L. H. S = \[\dfrac{{{{\text{a}}^3}{\text{ + }}{{\text{b}}^3}{\text{ + }}{{\text{c}}^3}}}{{3{\text{abc}}}}\]
As given in the question, \[{{\text{a}}^3}{\text{ }}{\text{ + }}{{\text{b}}^3}{\text{ + }}{{\text{c}}^3}{\text{ = 3abc}}\]. Applying this condition in the above equation
$ \Rightarrow $ L. H. S = \[\dfrac{{{\text{3abc}}}}{{3{\text{abc}}}}\] = 1 = R. H. S
Hence proved.
Note: To solve such types of questions, we have to follow a few steps. First, we will find the value of the variable which we will use to solve the problem. Then we will use the given condition in the question and put the value of the variable in the given condition to solve the given problem.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Trending doubts
Explain the Treaty of Vienna of 1815 class 10 social science CBSE

In cricket, what is the term for a bowler taking five wickets in an innings?

Who Won 36 Oscar Awards? Record Holder Revealed

What is the median of the first 10 natural numbers class 10 maths CBSE

Why is it 530 pm in india when it is 1200 afternoon class 10 social science CBSE

What is deficiency disease class 10 biology CBSE

