Question

give me the answer of this formulas please it is maths subject​

Answer 1

ANSWER:-

$1 ^{st} \: part$

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

As we know that any number with power 0 becomes

So

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

$2 ^{nd} part$

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

Here the power is only one so both the numbers will add respectively

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

$3 ^{rd} \: part$

$(a + b) ^{2} =$

Here the whole power is two so it will be

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

$4 ^{th} part$

$(a + b) {}^{3 = }$

Here the whole power is three , so it will be

$(a + b) {}^{3} = {a}^{3} + {b}^{3} + 3ab(a + b)$

$5 {}^{th} \: part$

$( {a - b})^{3} =$

Here the whole power is three so,it will be

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

$6^{th} \: part$

$(a + b + c) {}^{2} =$

Here the three digits have whole power 2 . So, the formula will be

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

If you don't understand refer the attachment given above

HOPE IT HELPS✌

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