Basic Maths: Test 43

If a +b +c= 0,
$\begin{align} & {{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \\ & Here,\,0.444+0.555+\left( -0.999 \right)=0 \\ & Therefore\,\,{{\left( 0.444 \right)}^{3}}+{{\left( 0.555 \right)}^{3}}+{{\left( -0.999 \right)}^{3}} \\ & =-3\times 0.444\times 0.555\times 0.999 \\ & Therefore\,\,\,\exp ression \\ & =\left[ -3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right)+3\left( 0.444 \right)\,\left( 0.555 \right)\left( 0.999 \right) \right]=0 \\ \end{align}$

$\begin{align} & Let\,\,\left( 2.53-4.07 \right)\,=a\,\,\left( 4.07-3.15 \right)=b\,\, \\ & and\,\,\left( 3.15-2.53 \right)=c\,\,Therefore\,\,a+b+c=0 \\ & Therefore\,\,\,\exp ression \\ & =\frac{{{a}^{2}}}{bc}+\frac{{{b}^{2}}}{ac}+\frac{{{c}^{2}}}{ab} \\ & =\frac{{{a}^{3}}+{{b}^{3}}+{{c}^{3}}}{abc}=\frac{3abc}{abc}=3 \\ & \left[ if\,\,a+b+c=0,\,{{a}^{3}}+{{b}^{3}}+{{c}^{3}}=3abc \right] \\ \end{align}$

$\begin{align} & =\frac{13}{15}\times \frac{13}{15}+\frac{3}{15}\times \frac{3}{15}-2\times \frac{13}{15}\times \frac{3}{15} \\ & ={{\left( \frac{13}{15}-\frac{3}{15} \right)}^{2}} \\ & ={{\left( \frac{13-3}{15} \right)}^{2}} \\ & ={{\left( \frac{10}{15} \right)}^{2}} \\ & =\frac{4}{9} \\ \end{align}$

Let 0.893= a
And 0.107= b
Therefore expression
$\begin{align} & =\frac{a\times a-b\times b}{a-b} \\ & =\frac{{{a}^{2}}-{{b}^{2}}}{a-b} \\ & =\frac{\left( a+b \right)\,\left( a-b \right)}{a-b} \\ & =a+b=0.893+0.107 \\ & =1 \\ \end{align}$

$\begin{align} & Let\,4\frac{5}{17}\,\,=a\,\,\,and\,\,\frac{17}{73}=b. \\ & Therefore\,\,\,\exp ression \\ & ={{\left( a+b \right)}^{2}}-{{\left( a-b \right)}^{2}} \\ & =\left( {{a}^{2}}+{{b}^{2}}+2ab \right)-\left( {{a}^{2}}+{{b}^{2}}-2ab \right) \\ & =4ab \\ & =4\times 4\frac{5}{17}\times \frac{17}{73} \\ & =4\times \frac{73}{17}\times \frac{17}{73} \\ & =4 \\ \end{align}$