Question

Show that the given equation is true.
$\left( a+\omega b+{{\omega }^{2}}c \right)\left( a+{{\omega }^{2}}b+\omega c \right)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab$

Answer 1

Hint:Firstly assume the LHS and simplify it by multiplying the brackets. After simplification arrange the similar terms to take out some common terms from them. Then use the properties \[{{\omega }^{3}}=1\] and \[\omega +{{\omega }^{2}}=-1\] to get the right hand side.

Complete step-by-step answer:
To prove the given equation is true we will write it down first,
$\left( a+\omega b+{{\omega }^{2}}c \right)\left( a+{{\omega }^{2}}b+\omega c \right)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab$
Now, we will first solve the left hand side of the equation and then therefore assume,
Left Hand Side (LHS) = $\left( a+\omega b+{{\omega }^{2}}c \right)\left( a+{{\omega }^{2}}b+\omega c \right)$ …………………………………. (1)
Right Hand Side (RHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab\] ……………………………………… (2)
Above equation can be simplified as,
Therefore, Left Hand Side (LHS) = $a\left( a+{{\omega }^{2}}b+\omega c \right)+\omega b\left( a+{{\omega }^{2}}b+\omega c \right)+{{\omega }^{2}}c\left( a+{{\omega }^{2}}b+\omega c \right)$
If we multiply inside the brackets in the above equation we will get,
Therefore, Left Hand Side (LHS) = $\left( {{a}^{2}}+{{\omega }^{2}}ab+\omega ac \right)+\left( \omega ab+{{\omega }^{3}}{{b}^{2}}+{{\omega }^{2}}bc \right)+\left( {{\omega }^{2}}ac+{{\omega }^{4}}bc+{{\omega }^{3}}{{c}^{2}} \right)$
If we open the brackets of the above equation we will get,
Therefore, Left Hand Side (LHS) = ${{a}^{2}}+{{\omega }^{2}}ab+\omega ac+\omega ab+{{\omega }^{3}}{{b}^{2}}+{{\omega }^{2}}bc+{{\omega }^{2}}ac+{{\omega }^{4}}bc+{{\omega }^{3}}{{c}^{2}}$
By rearranging the above equation we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{\omega }^{3}}{{b}^{2}}+{{\omega }^{3}}{{c}^{2}}+{{\omega }^{2}}ab+\omega ab+{{\omega }^{2}}bc+{{\omega }^{4}}bc+{{\omega }^{2}}ac+\omega ac\]
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{\omega }^{3}}{{b}^{2}}+{{\omega }^{3}}{{c}^{2}}+{{\omega }^{2}}ab+\omega ab+{{\omega }^{2}}bc+{{\omega }^{3+1}}bc+{{\omega }^{2}}ac+\omega ac\]
As we know that we can write \[{{a}^{m+n}}\] as \[{{a}^{m}}{{a}^{n}}\] therefore in the above equation we can replace \[{{\omega }^{3+1}}\] by \[{{\omega }^{3}}\omega \] therefore we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{\omega }^{3}}{{b}^{2}}+{{\omega }^{3}}{{c}^{2}}+{{\omega }^{2}}ab+\omega ab+{{\omega }^{2}}bc+{{\omega }^{3}}\omega bc+{{\omega }^{2}}ac+\omega ac\]
As we know that in the given equation \[1,\omega ,{{\omega }^{2}}\] are the cube roots of unity and to proceed further in the solution we should know the property of cube roots of unity given below,
Property:
\[{{\omega }^{3}}=1\]
If we use the above property in LHS we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+1\times {{b}^{2}}+1\times {{c}^{2}}+{{\omega }^{2}}ab+\omega ab+{{\omega }^{2}}bc+1\times \omega bc+{{\omega }^{2}}ac+\omega ac\]
Further simplification in the above equation will give,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+{{\omega }^{2}}ab+\omega ab+{{\omega }^{2}}bc+\omega bc+{{\omega }^{2}}ac+\omega ac\]
If we take ‘ab’ common from the above equation we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\left( {{\omega }^{2}}+\omega \right)ab+{{\omega }^{2}}bc+\omega bc+{{\omega }^{2}}ac+\omega ac\]
Similarly if we take ‘bc’ and ‘ac’ common from the above equation we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\left( {{\omega }^{2}}+\omega \right)ab+\left( {{\omega }^{2}}+\omega \right)bc+\left( {{\omega }^{2}}+\omega \right)ac\]
Now before we proceed further in the solution we should know the another important property of the cube roots of unity which is given below,
Property:
\[1+\omega +{{\omega }^{2}}=0\]
If we shift 1 on the right side of the equation we will get,
\[\omega +{{\omega }^{2}}=-1\]
If we use the above value in LHS we will get,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}+\left( -1 \right)ab+\left( -1 \right)bc+\left( -1 \right)ac\]
Further simplification in the above equation will give,
Therefore, Left Hand Side (LHS) = \[{{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab\]
If we compare the above equation with equation (2) we will get,
Left Hand Side (LHS) = Right Hand Side (RHS)
Therefore,
\[\left( a+\omega b+{{\omega }^{2}}c \right)\left( a+{{\omega }^{2}}b+\omega c \right)={{a}^{2}}+{{b}^{2}}+{{c}^{2}}-bc-ca-ab\]
Hence proved.
Therefore the given equation is true.

Note: Many students directly use the property \[1+\omega +{{\omega }^{2}}=0\] in the given equation without simplifying it and therefore face many difficulties in doing calculations. Do remember to simplify the equation so that you can use the simple value -1 of the property without any \[\omega \] to minimize your calculations.