Question

If there is given that four variables \[a,b,c,d\] are in continuous proportion, prove that \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)={{\left( ab+bc+cd \right)}^{2}}\].

Answer 1

Hint: In this question first we are going to choose which side from the given equation will be easy to solve and bring it equal to the opposite side but in this question as both the sides are totally different just by analyzing it then we will solve the question by solving both the sides to the extent that they are simplified to the end. It is given in the question that \[a,b,c,d\] are in continuous proportion which is \[\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\left( constant \right)\] and we will put the values of some variables in the form of other variables in the L.H.S. and R.H.S. of the question and will come to a common expression from both sides.

Complete step-by-step answer:
Now in these type of questions, from the given equation, we will first find that which side is easy to solve but in this question we can simply analyze that both sides will contain different terms when simplified, so in this situation we will solve both the sides of the equation to come to a common expression and prove both sides are equal.
Now in the question it is given, a very important detail or information, that \[a,b,c,d\] are in continuous proportion which means that these terms are in proportion and also continuously which means that in an perfect order which is \[\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=k\left( constant \right)\] but we need the first three only to solve the question, \[\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\left( constant \right)\].
By this above equation, \[\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\left( constant \right)\], we can deduce that,
\[\begin{align}
  & a=kb \\
 & b=kc \\
 & c=kd \\
\end{align}\]
First we will solve the L.H.S. of the question, which is,
\[L.H.S=\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]
Putting the above deduced values of \[a,b,c\] in the \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\] part of the above equation because we want to simplify the L.H.S. and make it equal to the R.H.S. so we can do it by making it a perfect square and so we are substituting in that specific part of L.H.S. only. We cannot put the deduced values in the \[\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\] part of the above equation because it will not become similar to \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\] and there will be no formation of a perfect square and we will not get close to the R.H.S. expression formation.
Now after putting the deduced values in the \[\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)\] part of the L.H.S. we get,
\[L.H.S.=\left[ {{\left( kb \right)}^{2}}+{{\left( kc \right)}^{2}}+{{\left( kd \right)}^{2}} \right]\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]
\[L.H.S.={{k}^{2}}\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)\]
\[L.H.S.={{k}^{2}}{{\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)}^{2}}\]
Now solving the R.H.S. of the question which is,
\[R.H.S.={{\left( ab+bc+cd \right)}^{2}}\]
Now putting the deduced values of variables in the R.H.S. we get,
\[R.H.S.={{\left[ \left( kb\cdot b \right)+\left( kc\cdot c \right)+\left( kd\cdot d \right) \right]}^{2}}\]
Solving further, we get,
\[R.H.S.={{\left( k{{b}^{2}}+k{{c}^{2}}+k{{d}^{2}} \right)}^{2}}\]
\[R.H.S.={{k}^{2}}{{\left( {{b}^{2}}+{{c}^{2}}+{{d}^{2}} \right)}^{2}}\]
Now as we can see that \[R.H.S.=L.H.S.\],
Hence proved.

Note: Do not give up on the question just by reading it, understand what is given and what is to be proven, write the relation given in the question and find some relation between the expressions or equations that are to be proven. Try to bring L.H.S. and R.H.S. as close as possible and when you can solve it no further then there might be some substitution or other concept given in the question, should be used. There is also an alternate method which involves deducing all the four variables and by putting the deduced values in the second bracket term of L.H.S. and also now we have to put different deducing variables in R.H.S.. other than before.