
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
600.9k+ 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.
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