{"id":6255,"date":"2021-12-21T07:52:40","date_gmt":"2021-12-21T07:52:40","guid":{"rendered":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/?page_id=6255"},"modified":"2021-12-23T11:45:13","modified_gmt":"2021-12-23T11:45:13","slug":"algebraic-identities-solved-examples","status":"publish","type":"page","link":"https:\/\/stgwebsite.mindspark.in\/studymaterial\/math-concepts\/algebraic-identities-solved-examples\/","title":{"rendered":"Algebraic Identities – Solved examples"},"content":{"rendered":"

[et_pb_section fb_built=”1″ admin_label=”Section” module_class=”mainsec” _builder_version=”4.10.4″ _module_preset=”default” background_color=”#e0f2fd” z_index=”1″ custom_padding=”5px||5px||true|false” locked=”off” collapsed=”off” global_colors_info=”{}”][et_pb_row column_structure=”3_5,2_5″ custom_padding_last_edited=”on|phone” _builder_version=”4.10.8″ _module_preset=”default” background_color=”#FFFFFF” width=”100%” max_width=”1310px” custom_padding=”|51px|40px|51px|false|true” custom_padding_tablet=”” custom_padding_phone=”|40px|30px|40px|false|true” border_radii=”on|10px|10px|10px|10px” global_colors_info=”{}”][et_pb_column type=”3_5″ admin_label=”Column L” _builder_version=”4.9.10″ _module_preset=”default” global_colors_info=”{}”][et_pb_text admin_label=”Acute Angles
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Algebraic Identities – Solved examples<\/h1>\n

[\/et_pb_text][et_pb_text admin_label=”Text” _builder_version=”4.13.1″ _module_preset=”default” text_font_size=”16px” header_2_font=”|600|||||||” header_2_text_color=”#a01414″ header_3_font=”|600|||||||” custom_padding=”15px|15px|54px|4px|false|false” hover_enabled=”0″ global_colors_info=”{}” sticky_enabled=”0″]<\/p>\n

Algebraic identities<\/b><\/h2>\n

The algebraic identities are specific types of algebraic expressions which are valid for any values of the variables present in them. These can be used for finding the values of variables, expansion or factorizing polynomials.<\/span><\/p>\n

There are some standard identities as given below. All other identities can be derived from these standard identities.<\/span><\/p>\n

 <\/p>\n

Standard Identities<\/b><\/h2>\n

1. (a+b)^{2}=a^{2}+b^{2}+2 a b<\/span>
2. (a-b)^{2}=a^{2}+b^{2}-2 a b<\/span>
3. (a+b)(a-b)=a^{2}-b^{2}<\/span>
4. (x+a)(x+b)=x^{2}+(a+b) x+a b<\/span>
5. (a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a<\/span>
6. (a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)<\/span>
7. (a-b)^{3}=a^{3}-b^{3}-3 a b(a-b)<\/span>
8. a^{3}+b^{3}+c^{3}=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+3 a b c<\/span><\/b><\/p>\n

\u00a0<\/b><\/p>\n

Proof of some basic identities (with the help of geometry)<\/b><\/h2>\n

We are going to prove the 3 most basic identities with the help of geometry.<\/span><\/p>\n

\\text { 1. }(a+b)^{2}=a^{2}+b^{2}+2 a b<\/span><\/span><\/p>\n

<\/span><\/p>\n

\"\"<\/span><\/p>\n

<\/span><\/p>\n

In the figure given above the square having side \u2018a + b\u2019 is divided into four parts.<\/span><\/p>\n

Area of the square having sides (a+b)=(a+b)=a^{2}+b^{2}+a b+a b<\/span><\/span><\/p>\n

\\Rightarrow(a+b)^{2}=a^{2}+b^{2}+2 a b<\/span><\/span><\/p>\n

<\/span><\/p>\n

\\text { 2. }(a-b)^{2}=a^{2}+b^{2}-2 a b<\/span><\/span><\/p>\n

\"\"<\/span><\/p>\n

In the figure given above the square having side \u2018a\u2019 is divided into three parts<\/span><\/p>\n

Area of the square having side \u2018a\u2019 =a^{2}=(a-b)^{2}+b(a-b)+a b<\/span><\/span><\/p>\n

\\Rightarrow a^{2}=(a-b)^{2}+a b-b^{2}+a b <\/span>
\\Rightarrow a^{2}=(a-b)^{2}-b^{2}+2 a b <\/span>
\\Rightarrow(a-b)^{2}=a^{2}+b^{2}-2 a b<\/span><\/p>\n

 <\/p>\n\\text { 3. }(x+a)(x+b)=x^{2}+(a+b) x+a b<\/span>\n

 <\/p>\n

\"\"<\/p>\n

In the figure given above the rectangle having length and breadth \u2018x+a\u2019 and \u2018x+b\u2019 respectively are divided into four parts<\/span><\/p>\n

Area of rectangle having sides (x+a) and (x+b)=(x+a)(x+b)=x^{2}+x a+x b+a b<\/span><\/span><\/p>\n

\\Rightarrow \\quad(x+a)(x+b)=x^{2}+(a+b) x+a b<\/span><\/span><\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

Solved Examples<\/b><\/h2>\n

 <\/p>\n\\text { 1) Prove that }(a+b)^{3}=a^{3}+b^{3}+3 a b(a+b)<\/span>\n

Solution<\/span><\/p>\n

\\text { First, we have to write }(a+b)^{3} \\text { as a product of }(a+b) \\text { and }(a+b)^{2}<\/span><\/span><\/p>\n

\\text { LHS }=(a+b)^{3}=(a+b)(a+b)^{2} <\/span>
=(a+b)\\left(a^{2}+b^{2}+2 a b\\right) <\/span>
=a\\left(a^{2}+b^{2}+2 a b\\right)+b\\left(a^{2}+b^{2}+2 a b\\right) <\/span>
=a^{3}+a b^{2}+2 a^{2} b+b a^{2}+b^{3}+2 a b^{2} <\/span>
=a^{3}+b^{3}+3 a^{2} b+3 a b^{2} <\/span>
=a^{3}+b^{3}+3 a b(a+b)=\\text { RHS }=\\text { Hence proved }<\/span>
<\/span><\/p>\n

 <\/p>\n

<\/span><\/p>\n

\\text { 2) Prove that }(a+b+c)^{2}=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a<\/span><\/span><\/p>\n

Solution:<\/span><\/p>\n

Let, a+b=x<\/span>
\\text { LHS }=(a+b+c)^{2}=(x+c)^{2} <\/span>
=x^{2}+c^{2}+2 x c <\/span>
=(a+b)^{2}+c^{2}+2(a+b) c <\/span>
=a^{2}+b^{2}+2 a b+c^{2}+2 c a+2 b c<\/span>
=a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a= RHS = Hence proved<\/span><\/span><\/p>\n

<\/span><\/p>\n

\\text { 3) Prove that }(a+b+c)^{3}=(a+b+c)\\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\\right)+3 a b c<\/span><\/span><\/p>\n

Solution<\/span><\/p>\n

(a+b+c)^{3}=(x+c)^{3}=x^{3}+c^{3}+3 x c(x+c) <\/span>
=x^{3}+c^{3}+3 x^{2} c+3 x c^{2}<\/span>
=(a+b)^{3}+c^{3}+3(a+b)^{2} c+3(a+b) c^{2} <\/span>
=a^{3}+b^{3}+3 a b(a+b)+c^{3}+3\\left(a^{2}+b^{2}+2 a b\\right) c+3(a+b) c^{2} <\/span>
=a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+3 a^{2} c+3 b^{2} c+6 a b c+3 a c^{2}+3 b c^{2} <\/span>
=a^{3}+b^{3}+c^{3}+3 a^{2} b+3 a b^{2}+3 a b c+3 b^{2} c+3 b c^{2}+3 a b c+3 a^{2} c+3 a b c+3 a c^{2}-3 a b c <\/span>
=a^{3}+b^{3}+c^{3}+3 a b(a+b+c)+3 b c(a+b+c)+3 c a(a+b+c)-3 a b c <\/span>
=a^{3}+b^{3}+c^{3}+3(a+b+c)(a b+b c+c a)-3 a b c<\/span><\/span><\/p>\n

 <\/p>\n

<\/span><\/p>\n

<\/span><\/p>\n

 <\/p>\n

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Explore Other Topics<\/h1>\n

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\n
Geometry<\/a><\/div>\n
Trigonometry<\/a><\/div>\n
Operations<\/a><\/div>\n
Numbers<\/a><\/div>\n<\/div>\n

[\/et_pb_text][et_pb_text admin_label=”Related Concepts
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Related Concepts<\/h1>\n

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\n
Prime numbers from 1 to 100<\/a><\/div>\n
53 prime or composite?<\/a><\/div>\n
Is 73 a prime number?<\/a><\/div>\n<\/div>\n

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Frequently Asked Questions\u00a0<\/span><\/span><\/h1>\n
    <\/ol>\n

    Q1. What are algebraic identities?
    <\/strong><\/h3>\n

    Ans: <\/strong><\/span>The algebraic identities are specific types of algebraic expressions which are valid for any values of the variables present in them. These can be used for finding the values of variables, expansion or factorizing polynomials.<\/span><\/p>\n

    Q2. How are algebraic expressions different from algebraic identities?
    <\/strong><\/h3>\n

    Ans: <\/strong>Algebraic expressions are valid for only some values of the variables but algebraic identities are valid for all values of the variables.

    <\/strong><\/p>\n

    [\/et_pb_text][\/et_pb_column][\/et_pb_row][\/et_pb_section]<\/p>\n","protected":false},"excerpt":{"rendered":"

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