Question

verify the following a=3 b=4
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Answer 1

Answer:

(3 + 4)² = 3²+ 2×3×4 +4² = 9 + 24 + 16 = 39

(3 - 4)² = 3²- 2×3×4 +4² = 9 - 24 +16 = 31

(3 + 4) (3 - 4) = 3² - 4² , 7×1 = 9 - 16 , 7 = 7 (LHS = RHS)

Answer 2

Answer:

Solution :

Verify the following for a = 3 and b = 4.

☼ a) (a + b)² = a² + 2ab + b²

Here

• ↝ a = 3
• ↝ b = 4

Now :-

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

$\longrightarrow\small\sf{(3 + 4)}^{2} = {3}^{2} + 2 \times 3 \times 4 + {4}^{2}$

${\longrightarrow{\small{\sf{(7)}^{2} = {(3 \times 3)} + 6 \times 4 + {(4 \times 4)}}}}$

${\longrightarrow{\small{\sf{(7 \times 7)} = 9 + 24 + 16}}}$

${\longrightarrow{\small{\sf{49 = 49}}}}$

$\longrightarrow{\small{\sf{\underline{\underline{LHS = RHS}}}}}$

Hence Verified!

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☼ b) (a - b)² = a² - 2ab + b²

Here :-

• ↝ a = 3
• ↝ b = 4

Now :-

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

$\longrightarrow\small\sf{(3 - 4)}^{2} = {3}^{2} - 2 \times 3 \times 4 + {4}^{2}$

$\longrightarrow\small\sf{( - 1)}^{2} = {(3 \times 3)} - 6\times 4 + {(4 \times 4)}$

${\longrightarrow{\small{\sf{( - 1 \times - 1)} = 9- 24+ 16}}}$

${\longrightarrow{\small{\sf{1} = 25- 24}}}$

${\longrightarrow{\small{\sf{1 = 1}}}}$

$\longrightarrow{\small{\sf{\underline{\underline{LHS = RHS}}}}}$

Hence Verified!

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☼ c) (a + b)(a - b) = a² - b²

Here :-

• ↝ a = 3
• ↝ b = 4

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

$\longrightarrow\small\sf{(3 + 4)(3- 4)= {3}^{2} - {4}^{2} }$

$\longrightarrow\small\sf{(7)( - 1)= (3 \times 3) - (4 \times 4) }$

$\longrightarrow\small\sf{7 \times - 1= 9 - 16 }$

$\longrightarrow\small\sf{-7= - 7}$

$\longrightarrow{\small{\sf{\underline{\underline{LHS = RHS}}}}}$

Hence Verified!

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Learn More :

☼ Algebraic identities:-

➛ (a + b)² + (a - b)² = 2a² + 2b²

➛ (a + b)² - (a - b)² = 4ab

➛ (a+b)(a -b) = a² - b²

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

➛ a² + b² = (a + b)² - 2ab

➛ (a + b)³ = a³ + b³ + 3ab ( a + b)

➛ (a - b)³ = a³ - b³ - 3ab ( a - b)

➛ a³ + b³ = (a +b)(a² - ab + b²)

➛ a³ - b³ = (a-b)(a² + ab +b²)

➛ If a + b + c = 0 then a³ + b³ + c³ = 3abc

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