It is given that
a, b, c are in continued proportion
So we get
a/b = b/c = k
(v) LHS = \[abc\text{ }{{\left( a\text{ }+\text{ }b\text{ }+\text{ }c \right)}^{3}}\]
We can write it as
\[=\text{ }c{{k}^{2}}.\text{ }ck.\text{ }c\text{ }{{[c{{k}^{2}}~+\text{ }ck\text{ }+\text{ }c]}^{3}}\]
Taking out the common terms
\[=\text{ }{{c}^{3}}~{{k}^{3}}~{{[c\text{ }({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)]}^{3}}\]
So we get
\[\begin{array}{*{35}{l}}
=\text{ }{{c}^{3}}~{{k}^{3}}.\text{ }{{c}^{3}}~{{({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)}^{3}} \\
=\text{ }{{c}^{6}}~{{k}^{3}}~{{({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)}^{3}} \\
\end{array}\]
RHS = \[~{{\left( ab\text{ }+\text{ }bc\text{ }+\text{ }ca \right)}^{3}}\]
We can write it as
= (ck2. ck + ck. c + c. ck2)3
So we get
\[\begin{array}{*{35}{l}}
=\text{ }{{({{c}^{2}}{{k}^{3}}~+\text{ }{{c}^{2}}k\text{ }+\text{ }{{c}^{2}}{{k}^{2}})}^{3}} \\
=\text{ }{{({{c}^{2}}{{k}^{3}}~+\text{ }{{c}^{2}}{{k}^{2}}~+\text{ }{{c}^{2}}k)}^{3}} \\
\end{array}\]
Taking out the common terms
\[\begin{array}{*{35}{l}}
=\text{ }{{[{{c}^{2}}k\text{ }({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)]}^{3}} \\
=\text{ }{{c}^{6}}{{k}^{3}}~{{({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)}^{3}} \\
\end{array}\]
Therefore, LHS = RHS.
(vi) LHS = (a + b + c) (a – b + c)
We can write it as
\[=\text{ }(c{{k}^{2}}~+\text{ }ck\text{ }+\text{ }c)\text{ }(c{{k}^{2}}~\text{ }ck\text{ }+\text{ }c)\]
Taking out the common terms
\[\begin{array}{*{35}{l}}
=\text{ }c\text{ }({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)\text{ }c\text{ }({{k}^{2}}~\text{ }k\text{ }+\text{ }1) \\
=\text{ }{{c}^{2}}~({{k}^{2}}~+\text{ }k\text{ }+\text{ }1)\text{ }({{k}^{2}}~\text{ }k\text{ }+\text{ }1) \\
\end{array}\]
So we get
\[=\text{ }{{c}^{2}}~({{k}^{4}}~+\text{ }{{k}^{2}}~+\text{ }1)\]
RHS = a2 + b2 + c2
We can write it as
\[=\text{ }{{(c{{k}^{2}})}^{2}}~+\text{ }{{\left( ck \right)}^{2}}~+\text{ }{{\left( c \right)}^{2}}\]
So we get
\[=\text{ }{{c}^{2}}{{k}^{4}}~+\text{ }{{c}^{2}}{{k}^{2}}~+\text{ }{{c}^{2}}\]
Taking out the common terms
\[=\text{ }{{c}^{2}}~({{k}^{4}}~+\text{ }{{k}^{2}}~+\text{ }1)\]
Therefore, LHS = RHS.