Question

If -b=4 and ab=21 , find the value of a^3-b^3.

Answer 1

${ \large{ \underbrace{ \boxed{ \mathfrak{ \color{deeppink} {Answer }}}}}}$

${\rightarrow{ \boxed{ \mathfrak{ \color{lime} { {a}^{3} - {b}^{3} = 316}}}}}$

${ \large{ \underbrace{ \boxed{ \mathfrak{ \color{aqua} {Explanation }}}}}}$

☆To find :-

a³-b³

☆Given :-

a -b = 4 and ab = 21

☆Formula :-

${\rightarrow{ \boxed{ \mathfrak{ \color{yellow} { {a}^{3} - {b}^{3} = ( {a - b})^3 + 3ab(a - b)}}}}}$

☆Solution :-

Now putting the values in the formula

$\sf \: {a}^{3} - {b}^{3} \Rightarrow({4})^{3} + 3 \times 21 \times 4 \\ \\ \Rightarrow64 + 252 \: \\ \\ \Rightarrow \mathfrak \color{red}316 \qquad \:$

☆Hence :-

a³-b³=316

☆Derivation of formula :-

We know, (a-b)³ =a³-3a²b+3ab²-b³

If we subtract 3ab(a-b) i.e.,3a²b-3ab² from

(a-b)³ i.e.,a³-3a²b +3ab²-b³ we will get a³-b³

By solving (a-b)³+3ab(a-b)

= a³-3a²b+3ab²-b²+3a²b-3ab²

=a³-b³+3a²b-3a²b+3ab²-3ab²

=a³- b³

We get a³-b³ as (a-b)³+3ab(a-b)

There are many more formulas but for this type of question this formula will be used .

As the values of (a-b) and ab is given.

___________________________

☆Some algebraic formulas :-

• a² - b² = (a+b)(a-b)

• (a+b)² = a²+2ab +b²

• (a-b)² = a²-2ab +b²

• (a + b + c)²= a² + b²+ c²+ 2ab + 2bc + 2ca.

• (a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ca.

• (a + b)³ = a³ + 3a²b + 3ab² + b³

• (a + b)³ = a³ + b³+ 3ab(a + b)

€€________________________€€

hope it helps..... :)