Question

The value of (997)1/3 according to Binomial Theorem

Answer 1

The binomial theorem states the expression of a long term turned short through the convention of the formula as  where      

$(a+b)^{ n }=\Sigma ^{ n }_{ k=0 }(nk)a^{ n-k }b^{ k }(nk)=^{ n }C_{ k }=\frac { n! }{ (n-k)!k! }$

$Therefore,given\quad term\quad ({ 997) }^{ \frac { 1 }{ 3 } }can\quad be\quad written\quad as\quad { (1000-3) }^{ \frac { 1 }{ 3 } }$

$\Rightarrow { (1000-3) }^{ \frac { 1 }{ 3 } }=\quad { (1000) }^{ \frac { 1 }{ 3 } }{ \left( 1-\frac { 3 }{ 1000 } \right) }^{ \frac { 1 }{ 3 } }\quad =\quad 10{ \left( 1-0.003 \right) }^{ \frac { 1 }{ 3 } }$

So,if $\quad x<<<1\quad then\quad { \left( 1-x \right) }^{ n }\quad becomes\quad (1-nx),$

Therefore, $\quad \Rightarrow { 10(1-0.003) }^{ \frac { 1 }{ 3 } }$

$=10(1 \times \frac { 1 }{ 3 } - 0.003 \times \frac { 1 }{ 3 } )$

$=10(\frac { 1 }{ 3 } -0.001)$

$=10(0.333-0.001)=3.323$