Question

find the value of 4^0 -3^0 into 6^0​

Answer 1

We have been asked to find the value of: $\tt{ {(4)}^{0} - {(3)}^{0} \times {(6)}^{0}}$. In order to find the same we have to recall the laws of indices. In these laws of indices ; there is a law which says :

$\large{\boxed{\tt{ {(a)}^{0} = 1}}}$

Here (a) can be any number. When any number is raised to zero ; the value or product is always one. Reason being ; zero means no power , and when a number has no power it gives zero.

So , following the above information we will solve the given problem.

$\longrightarrow{\tt{{(4)}^{0} - {(3)}^{0} \times {(6)}^{0} }}$

Keeping the law in mind ; we get the values as 1.

$\longrightarrow{\tt{ 1 - 1 \times 1}}$

Further simplification gives :

$\longrightarrow{\tt{0 \times 1}}$

Finally we get :

$\large{\boxed{\tt{0}}}$

Answer 2

$\huge \bf \orange { Question }$

Find the Value of $\bf {4}^{0} - {3}^{0} \times {6}^{0}$

$\huge \bf \purple{Answer}$

We know the law of Indices that is $\bf {x}^{0} = 1$

This law States that if any Number is raised to the Power Zero, it's value is 1.

Eg. → $\sf {1}^{0} = 1$

$\sf {(-8)}^{0} = 1$

$\sf {( \dfrac{2}{4} ) }^{0} = 1$

Thus,

$\longrightarrow \bf {(4)}^{0} - {(3)}^{0} \times {(6)}^{0}$

${\longrightarrow \bf 1 - 1 \times 1}$

Equating it Further on BODMAS rule

$\red{ \implies \bf1 - 1}$

$\red{ \implies \bf \boxed{ \boxed {0}}}$

Thus, we got the Final Answer 0

Additional Information

$\sf {(a+b)^{0} = 1}$

$\sf {(a+b)^{1} = a + b}$

$\sf {(a+b)^{2} = a^{2} + 2ab + b^{2}}$

$\sf {(a-b)^{2} = a^{2} - 2ab + b^{2}}$

$\sf { a^{2} - b^{2} = (a+b)(a-b) }$

$\sf { {a}^{2} + {b}^{2} = (a + b {)}^{2} - 2ab }$

$\sf { {a}^{2} + {b}^{2} = (a - b {)}^{2} + 2ab }$

$\sf { {a}^{3} + {b}^{3} = (a + b)( {a}^{2} - ab + {b}^{2}) }$

$\sf { {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2}) }$

$\sf {(a + b + c {)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca}$

$\sf { {a}^{3} + {b}^{3} + {c}^{3} = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca) }$