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If a, b, c are in continued proportion, prove that: (v)

    \[\mathbf{abc}\text{ }{{\left( \mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)}^{\mathbf{3}}}~=\text{ }{{\left( \mathbf{ab}\text{ }+\text{ }\mathbf{bc}\text{ }+\text{ }\mathbf{ca} \right)}^{\mathbf{3}}}\]

(vi)

    \[\left( \mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)\text{ }=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}}\]

Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If a, b, c are in continued proportion, prove that: (v)

    \[\mathbf{abc}\text{ }{{\left( \mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)}^{\mathbf{3}}}~=\text{ }{{\left( \mathbf{ab}\text{ }+\text{ }\mathbf{bc}\text{ }+\text{ }\mathbf{ca} \right)}^{\mathbf{3}}}\]

(vi)

    \[\left( \mathbf{a}\text{ }+\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)\text{ }\left( \mathbf{a}\text{ }\text{ }\mathbf{b}\text{ }+\text{ }\mathbf{c} \right)\text{ }=\text{ }{{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}}\]

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.

i

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