Question

Formula of (a+b)3 and (a-b)3?​

Answer 1

Step-by-step explanation:

$\blue{ \bold{ \underline{ QUESTION : - }}}$

Formula of (a+b)³ and (a-b)³ ?

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$\star{ \pink{ \underline{ \underline{ \bold{ReQuRiEd \: \: AnSWeR : - }}}}}$

• (a+b)³ = a³+b³ +3ab(a+b)

Or. (a+b)³ = a³ +b³ +3a²b+3ab²

• (a-b)³ = a³-b³-3ab(a-b)

Or. (a-b)³ = a³ -b³-3a²b+ 3ab²

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$\boxed{ \huge{\bold{ \red{ \underbrace{ Verification}}}}}$

LET,

• a = 2

• b = 4

(a-b)³ =a³ -b³-3a²b+ 3ab²

=> (2-4)³ = 2³-4³-3×2²×4+ 3× 2× 4²

=>(- 2)³ = 8 - 48 + 96 -64

=> -8 = -8

$\boxed{ \green{ \mathfrak{L.H.S = R.H.S}}}$

Agian,

• a = 2

• b = 4

(a+b)³ = a³ +b³ +3a²b+3ab²

=> (2+4)³ = 2³+4³+3×2²×4 + 3×2× 4²

=> (6)³ = 8 +48 + 96 +64

=> 216 = 216

$\boxed{ \green{ \mathfrak{L.H.S = R.H.S}}}$

We see Both cases Are L.H.S = R.H.S so The Both

Formula Are Proved

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MORE YOU KNOW ➡️

• a²– b² = (a – b) (a + b)

• (a + b)² = a² + 2ab + b²

• a² + b² = (a + b)² – 2ab

• (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

• a³ – b³ = (a – b) (a²+ ab + b²)

• a³+ b³ = (a + b) (a²– ab + b²)

• (a + b)⁴ = a⁴ + 4a³b + 6a²b² + 4ab³ + b⁴

• (a – b)⁴= a⁴– 4a³b + 6a²b² – 4ab³ + b⁴

• a⁴– b⁴ = (a – b) (a + b) (a² + b²)

• a⁵ – b⁵ = (a – b) (a⁴ + a³b + a2b² + ab³ + b⁴)

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Answer 2

$\Huge{\color{magenta}{Solution:-}}$

(a+b)³ = a³+b³ +3ab(a+b)

(a-b)³ = a³-b³-3ab(a-b)

_________________________

$\Huge{\color{magenta}{Verification:-}}$

LET,

a = 2

b = 4

(a-b)³ =a³ -b³-3a²b+ 3ab²

$\longrightarrow$ (2-4)³ = 2³-4³-3×2²×4+ 3× 2× 4²

$\longrightarrow$(- 2)³ = 8 - 48 + 96 -64

$\longrightarrow$ -8 = -8

$\boxed{ \red{ \mathtt{L.H.S = R.H.S}}}$

Again Let,

a = 2

b = 4

(a+b)³ = a³ +b³ +3a²b+3ab²

$\longrightarrow$ (2+4)³ = 2³+4³+3×2²×4 + 3×2× 4²

$\longrightarrow$ (6)³ = 8 +48 + 96 +64

$\longrightarrow$216 = 216

$\boxed{ \blue{ \mathtt{L.H.S = R.H.S}}}$