Question

Prove that: \[{\left( {\dfrac{{{3^a}}}{{{3^b}}}} \right)^{a + b}}{\left( {\dfrac{{{3^b}}}{{{3^c}}}} \right)^{b + c}}{\left( {\dfrac{{{3^c}}}{{{3^a}}}} \right)^{c + a}} = 1\].

Answer 1

Hint: The given problem revolves around the concepts of algebraic solution, which entirely depends on applying the certain rules of indices. Solving the L.H.S. can automatically prove the R.H.S. Since, the bases are already equal i.e. same in L.H.S. part particularly, As a result, directly applying the rules of indices to the left hand side (L.H.S.) of the given equation (also, the solution will be in need of expansion formulae such as \[{\left( {a + b} \right)^2}\] respectively), the desired conclusion is obtained.

Complete step-by-step solution:
Since, we have given the equation that,
\[{\left( {\dfrac{{{3^a}}}{{{3^b}}}} \right)^{a + b}}{\left( {\dfrac{{{3^b}}}{{{3^c}}}} \right)^{b + c}}{\left( {\dfrac{{{3^c}}}{{{3^a}}}} \right)^{c + a}} = 1\]
As a result, we have to prove the ‘\[{\text{L}}{\text{.H}}{\text{.S}}{\text{. = R}}{\text{.H}}{\text{.S}}{\text{.}}\]’ as asked in the question
Therefore, let us solve the L.H.S. to prove R.H.S. i.e.
\[L.H.S. = {\left( {\dfrac{{{3^a}}}{{{3^b}}}} \right)^{a + b}}{\left( {\dfrac{{{3^b}}}{{{3^c}}}} \right)^{b + c}}{\left( {\dfrac{{{3^c}}}{{{3^a}}}} \right)^{c + a}}\]
Since, all the bases are same (equal) so using the certain rules of indices the desired equation can be proved,
Hence, now, since taking the all of the denominators (present), in the numerator, where the indices of the certain terms becomes inverse of its respective sign such as positive becomes negative and negative becomes positive respectively (according to the mathematics) [also, this is one of the rules of indices for dividation i.e. \[\dfrac{{{a^x}}}{{{a^y}}} = {a^{x - y}}\]], we get
\[L.H.S. = {\left( {{3^{a - b}}} \right)^{a + b}}{\left( {{3^{b - c}}} \right)^{b + c}}{\left( {{3^{c - a}}} \right)^{c + a}}\]
Hence, now by using the rules for indices that is , \[{\left( {{a^x}} \right)^y} = {a^{xy}}\] respectively, we get
\[L.H.S. = {\left( 3 \right)^{\left( {a - b} \right)\left( {a + b} \right)}}{\left( 3 \right)^{\left( {b - c} \right)\left( {b + c} \right)}}{\left( 3 \right)^{\left( {c - a} \right)\left( {c + a} \right)}}\]
Now, we know that the expansion formulae for \[\left( {x - y} \right)\left( {x + y} \right)\] is \[\left( {{x^2} - {y^2}} \right)\], we get
 \[L.H.S. = {\left( 3 \right)^{\left( {{a^2} - {b^2}} \right)}}{\left( 3 \right)^{\left( {{b^2} - {c^2}} \right)}}{\left( 3 \right)^{\left( {{c^2} - {a^2}} \right)}}\]
Hence, again using the certain rules for indices such as, \[{\left( a \right)^x} \times {\left( a \right)^y} = {\left( a \right)^{x + y}}\] (multiplication rule in indices), we get
\[L.H.S. = {\left( 3 \right)^{{a^2} - {b^2} + {b^2} - {c^2} + {c^2} - {a^2}}}\]
Solving the equation mathematically, we get
\[L.H.S. = {\left( 3 \right)^0}\]
Where, rules of indices say that \[{a^0} = 1\], we get
\[L.H.S. = 1\]
\[\therefore L.H.S. = R.H.S.\]
Hence, proved!
\[\therefore \Rightarrow \] The option (c) is absolutely correct.

Note: While applying the rules of certain indices remember that the bases should be the same. One must also know all the rules of indices when the certain terms are (especially) in multiplication, dividation stage such as \[{a^m} \times {a^n} = {a^{m + n}}\], \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\], \[{\left( {{a^m}} \right)^n} = {a^{mn}}\], etc. (also, discussed briefly) which seems the most important thing in solving any algebraic solutions to get the desired output. Also, remember \[{a^m} = \dfrac{1}{{{a^{ - m}}}}\], \[{\text{anything }^0} = 1\], \[\sqrt[n]{a} = {a^{\dfrac{1}{n}}}\]. Empathize the knowledge of algebraic terms which includes the variables x, y, z,… respectively, which seems the complexity of the respective problem particularly by knowing certain type of expansion formulae i.e. \[{\left( {a + b} \right)^2}\], \[{\left( {a - b} \right)^2}\], \[{\left( {a + b} \right)^3}\], etc., so as to be sure of our final answer.