Question

expand:(2x+1)³ answer ​

Answer 1

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❐QuEsTiOn,

$\pink{ \sf \: Expand:(2x+1)³ answer }$

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❐SoLuTiOn,

$\green{\sf {↪(2x + 1)}^{3} = {(2x + 1)}^{3} }$

$\pink{ \sf \underline{\: As \: We \: know \: That, }}$

• $\green{\sf \: \boxed{ \pink{{(a + b)}^{3} = {a}^{3} + 3ab(a + b) + {b}^{3} }}}$

$\green{ \sf \underline{\: Then, }}$

$\pink{\sf \: ↪ {(2x + 1)}^{3} = {(2x + 1)}^{3} }$

$\green{ \sf \: ↪ {(2x + 1)}^{3} = {(2x)}^{3} + 3(2x)(1)(2x + 1) + {1}^{3} }$

$\pink{ \sf \: ↪ {(2x + 1)}^{3} = {8x}^{3} + 6x(2x + 1) + 1 }$

$\green{ \sf \: ↪ {(2x + 1)}^{3} = {8x}^{3} + 12 {x}^{2} + 6x + 1 }$

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❐FiNal AnSwEr,

$\blue{ \sf \: {(2x + 1)}^{3} = {8x}^{3} + 12 {x}^{2} + 6x + 1 }$

❥Hope you get your AnSwEr.

Answer 2

$\large \underline{ \underline{ \text{Question:}}}$

• $\text{Expand: } \: {(2x + 1)}^{3}$

$\large \underline{ \underline{ \text{Solution:}}}$

$\longrightarrow {(2x + 1)}^{3}$

$[ {(x + y)}^{3} = {x}^{3} + 3 {x}^{2}y + 3x {y}^{2} + {y}^{3} ]$

$\longrightarrow {(2x + 1)}^{3}$

$\longrightarrow {(2x)}^{3} + 3{(2x)}^{2}(1) + 3(2x)(1)^{2} + {1}^{3}$

$\longrightarrow {8x}^{3} + 3{(4 {x}^{2} )}(1) + 3(2x)(1) + {1}$

$\longrightarrow \boxed{{8x}^{3} + 12 {x}^{2} + 6x + 1}$

$\large \underline{ \underline{ \text{Identity Used:}}}$

• ${(x + y)}^{3} = {x}^{3} + 3 {x}^{2}y + 3x {y}^{2} + {y}^{3}$

$\large \underline{ \underline{ \text{Final Answer:}}}$

• $\text{The expansion of } {(2x + 1)}^{3} \: \text{is} \: \: \: \: \: \: \: \: \: \: \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \ 8 {x}^{3} + 12 {x}^{2} + 6x + 1.$

$\large \underline{ \underline{ \text{More Information:}}}$

$\large \underline{{ \text{More Algebraic Identities,}}}$

• ${(x + y)}^{2} = {x}^{2} + 2xy + {y}^{2}$
• ${(x - y)}^{2} = {x}^{2} - 2xy + {y}^{2}$
• $(x + y)(x - y) = {x}^{2} - {y}^{2}$
• $(x + a)(x + b) = {x}^{2} + (a + b)x + ab$
• ${(x + y + z)}^{2} = {x}^{2} + {y}^{2} + {z}^{2} + 2xy + 2yz + 2zx$
• ${(x - y)}^{3} = {x}^{3} + 3 {x}^{2} y - 3x {y}^{2} - {y}^{3}$
• ${x}^{3} + {y}^{3} + {z}^{3} - 3xyz = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx)$