Chapter 2: Polynomials

Ans: We have

 $ {\left( {3a - 2b} \right)^3} = {\left( {3a} \right)^3} - {\left( {2b} \right)^3} - 3\left( {3a} \right)\left( {2b} \right)\left( {3a - 2b} \right) $ 

Now, using identity  $ \left[ {{{\left( {a - b} \right)}^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)} \right] $ 

 $  = 27{a^3} - 8{b^3} - 54{a^2}b + 36a{b^2} $ 

Ans: Here, using the identity  $ \left[ {{{\left( {a + b} \right)}^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)} \right] $ 

 $ {\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right)^3} = {\left( {\dfrac{1}{x}} \right)^3} + {\left( {\dfrac{y}{3}} \right)^3} + 3 \times \dfrac{1}{x} \times \dfrac{y}{3}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $ 

 $  \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{x}\left( {\dfrac{1}{x} + \dfrac{y}{3}} \right) $ 

 $  \Rightarrow \dfrac{1}{{{x^3}}} + \dfrac{{{y^3}}}{{27}} + \dfrac{y}{{{x^2}}} + \dfrac{{{y^2}}}{{3x}} $ 

Ans: We have

 $ {\left( {4 - \dfrac{1}{{3x}}} \right)^3} = {\left( 4 \right)^3} - {\left( {\dfrac{1}{{3x}}} \right)^3} - 3\left( 4 \right)\left( {\dfrac{1}{{3x}}} \right)\left( {4 - \dfrac{1}{{3x}}} \right) $ 

Now, using identity  $ \left[ {{{\left( {a - b} \right)}^3} = {a^3} - {b^3} - 3ab\left( {a - b} \right)} \right] $ 

 $  = 64 - \dfrac{1}{{27{x^3}}} - \dfrac{4}{x}\left( {4 - \dfrac{1}{{3x}}} \right) $ 

 $  = 64 - \dfrac{1}{{27{x^3}}} - \dfrac{{16}}{x} + \dfrac{4}{{3{x^2}}} $ 

Ans: We have,

 $ 1 - 64{a^3} - 12a + 48{a^2} = {\left( 1 \right)^3} - {\left( {4a} \right)^3} - 3\left( 1 \right)\left( {4a} \right)\left( {1 - 4a} \right) $ 

 $  = {\left( {1 - 4a} \right)^3}\left[ {{a^3} - {b^3} - 3ab\left( {a - b} \right) = {{\left( {a - b} \right)}^3}} \right] $ 

 $  = \left( {1 - 4a} \right)\left( {1 - 4a} \right)\left( {1 - 4a} \right) $ 

Ans: We have,

$ 8{p^3} + \dfrac{{12}}{5}{p^2} + \dfrac{6}{{25}}p + \dfrac{1}{{125}} $ 

 $  = {\left( {2p} \right)^3} + 3 \times {\left( {2p} \right)^2} \times \dfrac{1}{5} + 3 \times \left( {2p} \right) \times {\left( {\dfrac{1}{5}} \right)^2} + {\left( {\dfrac{1}{5}} \right)^3} $ 

 $  = {\left( {2p} \right)^3} + {\left( {\dfrac{1}{5}} \right)^3} + 3 \times \left( {2p} \right) \times \dfrac{1}{5}\left[ {2p + \dfrac{1}{5}} \right] $ 

 $  = {\left( {2p + \dfrac{1}{5}} \right)^3} = \left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right)\left( {2p + \dfrac{1}{5}} \right) $ 

Ans: We have,

 $ \left( {\dfrac{x}{2} + 2y} \right)\left( {\dfrac{{{x^2}}}{4} - xy + 4{y^2}} \right) = \left( {\dfrac{x}{y} + 2y} \right)\left( {{{\left( {\dfrac{x}{2}} \right)}^2} - \left( {\dfrac{x}{2}} \right)\left( {2y} \right) + {{\left( {2y} \right)}^2}} \right) $ 

 $  = {\left( {\dfrac{x}{2}} \right)^3} + {\left( {2y} \right)^3}\left[ {\left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) = {a^3} + {b^3}} \right] $ 

 $  = \dfrac{{{x^3}}}{8} + 8{y^3} $ 

Ans: We have,

 $ \left( {{{\left( x \right)}^2} - 1} \right)\left( {{x^4} + {x^2} + 1} \right) = \left( {{x^2} - 1} \right)\left( {{{\left( {{x^2}} \right)}^2} + \left( {{x^2}} \right)\left( 1 \right) + {{\left( 1 \right)}^2}} \right) $ 

 $  = {\left( {{x^2}} \right)^3} - {\left( 1 \right)^3}\left[ {\left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) = {a^3} - {b^3}} \right] $ 

 $  = {x^6} - 1 $ 

Ans: We have,

 $ 1 + 64{x^3} = {\left( 1 \right)^3} + {\left( {4x} \right)^3} $ 

 $  = \left( {1 + 4x} \right)\left( {{{\left( 1 \right)}^2} - \left( 1 \right)\left( {4x} \right) + {{\left( {4x} \right)}^2}} \right) $ 

 $  = \left( {1 + 4x} \right)\left( {1 - 4x + 16{x^2}} \right) $ 

Ans: We have,

 $ {a^3} - 2\sqrt {2{b^3}}  = {\left( a \right)^3} - {\left( {\sqrt 2 b} \right)^3} $ 

 $  = \left( {a - \sqrt 2 b} \right)\left( {{{\left( a \right)}^2} + \left( a \right)\left( {\sqrt 2 b} \right) + {{\left( {\sqrt 2 b} \right)}^2}} \right) $ 

 $  = \left( {a - \sqrt 2 b} \right)\left( {{a^2} + \sqrt 2 ab + 2{b^2}} \right) $ 

Ans: We have,

 $ \left( {2x - y + 3z} \right)\left( {4{x^2} + {y^2} + 9{z^2} + 2xy + 3yz - 6xz} \right) $ 

 $  = \left\{ {2x + \left( { - y} \right) + 3z} \right\}\left\{ {{{\left( {2x} \right)}^2} + {{\left( { - y} \right)}^2} + {{\left( {3z} \right)}^2} - 2x\left( { - y} \right) - \left( { - y} \right)\left( {3z} \right) - \left( { - y} \right)\left( {3z} \right) - \left( {3z} \right)\left( {2x} \right)} \right\} $ 

 $  = {\left( {2x} \right)^3} + {\left( { - y} \right)^3} + {\left( {3z} \right)^3} - 3\left( {2x} \right)\left( { - y} \right)\left( {3z} \right) $  

using identity  $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  = 8{x^3} - {y^3} + 27{z^2} + 18xyz $ 

Ans: We have,

 $ {a^3} - 8{b^3} - 64{c^3} - 24abc $ 

 $  = \left\{ {{{\left( a \right)}^3} + {{\left( { - 2b} \right)}^3} + {{\left( { - 4c} \right)}^3} - 3\left( a \right)\left( { - 2b} \right)\left( { - 4c} \right)} \right\} $ 

 $  = \left\{ {a + \left( { - 2b} \right) + \left( { - 4c} \right)} \right\}\left\{ {{a^2} + {{\left( { - 2b} \right)}^2} + {{\left( { - 4c} \right)}^2} - a\left( { - 2b} \right) - \left( { - 2b} \right)\left( { - 4c} \right) - \left( { - 4c} \right)a} \right\} $ 

Using identity  $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  = \left( {a - 2b - 4c} \right)\left( {{a^2} + 4{b^2} + 16{c^2} + 2ab - 8bc + 4ca} \right) $ 

Ans: We have,

 $ 2\sqrt 2 {a^3} + 8{b^3} - 27{c^3} + 18\sqrt 2 abc $ 

 $  = \left\{ {{{\left( {\sqrt 2 a} \right)}^3} + {{\left( {2b} \right)}^3} + {{\left( { - 3c} \right)}^3} - 3\left( {\sqrt 2 a} \right)\left( {2b} \right)\left( { - 3c} \right)} \right\} $ 

 $  = \left\{ {\sqrt 2 a + 2b + \left( { - 3c} \right)} \right\}\left\{ {{{\left( {\sqrt 2 a} \right)}^2} + {{\left( {2b} \right)}^2} + {{\left( { - 3c} \right)}^2} - \left( {\sqrt 2 a} \right)\left( {2b} \right) - \left( {2b} \right)\left( { - 3c} \right) - \left( { - 3c} \right)\left( {\sqrt 2 a} \right)} \right\} $ 

Using identity  $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

 $  = \left( {\sqrt 2 a + 2b - 3c} \right)\left( {2{a^2} + 4{b^2} + 9{c^2} - 2\sqrt 2 ab + 6bc + 3\sqrt 2 ca} \right) $ 

Ans:  Let  $ a = \dfrac{1}{2}\,,\,b = \dfrac{1}{3}\,,\,c =  - \dfrac{5}{6} $ 

 $ a + b + c = \dfrac{1}{2} + \dfrac{1}{3} - \dfrac{5}{6} $ 

 $  = \dfrac{{3 + 2 - 5}}{6} = \dfrac{0}{6} = 0 $ 

 $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

Therefore, 

 $  \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $ 

 $  \Rightarrow 3 \times \dfrac{1}{2} \times \dfrac{1}{3} \times  - \dfrac{5}{6} $ 

 $  \Rightarrow  - \dfrac{5}{{12}} $ 

Ans: We have,

 $ {\left( {0.2} \right)^3} - {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} = {\left( {0.2} \right)^3} + {\left( { - 0.3} \right)^3} + {\left( {0.1} \right)^3} $ 

 $ a = 0.2\,,\,b =  - 0.3\,,\,c = 0.1 $ . Then,

 $ a + b + c = 0.2 - 0.3 - 0.1 = 0 $ .

 $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

Therefore, 

 $  \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $ 

 $  \Rightarrow 3 \times 0.2 \times \left( { - 0.3} \right) \times \left( {0.1} \right) $ 

 $  \Rightarrow  - 0.018 $ 

Hence,  $ {\left( {0.2} \right)^3} - {\left( {0.3} \right)^3} + {\left( {0.1} \right)^3} =  - 0.018 $ 

Ans: We have,

 $ {\left( {x - 2y} \right)^3} + {\left( {2y - 3z} \right)^3} + {\left( {3z - x} \right)^3} $ 

 $ a = x - 2y\,,\,b = 2y - 3z\,,\,c = 3z - x $ . Then,

 $ a + b + c = x - 2y + 2y - 3z + 3z - x = 0 $ .

 $ \left[ {\left( {a + b + c} \right)\left( {{a^2} + {b^2} + {c^2} - ab - bc - ca} \right) = {a^3} + {b^3} + {c^3} - 3abc} \right] $ 

Therefore, 

 $  \Rightarrow {a^3} + {b^3} + {c^3} = 3abc $ 

 $  \Rightarrow 3 \times \left( {x - 2y} \right) \times \left( {2y - 3z} \right) \times \left( {3z - x} \right) $ 

 $  \Rightarrow 3\left( {x - 2y} \right)\left( {2y - 3z} \right)\left( {3z - x} \right) $ 

Ans: $ {x^3} + {y^3} - 12xy + 64 = {x^3} + {y^3} + {4^3} - 3xy\left( 4 \right) $ 

 $  = \left( {x + y + 4} \right)\left( {{x^2} + {y^2} + {4^2} - xy - 4y - 4x} \right) $ 

 $ \left[ {x + y =  - 4} \right] $ 

 $  = \left( 0 \right)\left( {{x^2} + {y^2} + {4^2} - xy - 4y - 4x} \right) = 0 $ 

Ans: $ {x^3} - 8{y^3} - 36xy - 216 = {x^3} + {\left( { - 2y} \right)^3} + {\left( { - 6} \right)^3} - 3x\left( { - 2y} \right)\left( { - 6} \right) $ 

 $  = \left( {x - 2y - 6} \right)\left( {{x^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - 6} \right)}^2} - x\left( { - 2y} \right) - \left( { - 2y} \right)\left( { - 6} \right) - x\left( { - 6} \right)} \right) $ 

 $  = \left( {x - \left( {2y + 6} \right)} \right)\left( {{x^2} + {{\left( { - 2y} \right)}^2} + {{\left( { - 6} \right)}^2} - x\left( { - 2y} \right) - \left( { - 2y} \right)\left( { - 6} \right) - x\left( { - 6} \right)} \right) $ 

 $ \left[ {x = 2y + 6} \right] $ 

 $  = \left( 0 \right)\left( {{x^2} + 4{y^2} + 36 + 2xy - 12y + 6x} \right) = 0 $ 

Ans: Area  $ 4{a^2} + 4a - 3 $ .

Here we will use a method of splitting the middle term.

Here we have to split the middle term as  $ 4a = 6a - 2a $ 

Now,

  $  \Rightarrow 4{a^2} + \left( {6a - 2a} \right) - 3 $ 

 $  \Rightarrow 4{a^2} + 6a - 2a - 3 $ 

 $  \Rightarrow 2a\left( {2a + 3} \right) - 1\left( {2a + 3} \right) $ 

 $  \Rightarrow \left( {2a - 1} \right)\left( {2a + 3} \right) $ 

We know that, area of rectangle \[ = 4{a^2} + 4a - 3\]

Here we also know that, area of a rectangle  $  = \,length\, \times \,breadth $  and  $ 4{a^2} + 4a - 3 = \left( {2a - 1} \right)\left( {2a + 3} \right) $ .

Therefore, its possible length and breadth  $  = \,\left( {2a - 1} \right) $  and   $ \left( {2a + 3} \right) $  or, we can say that  $ length = \left( {2a + 3} \right) $  and  $ breadth\, = \,\left( {2a - 1} \right) $ .