Question

Solve it................​

Answer 1

Solution:-

24) take 3x^7 as common

=3x^7(x^6-64y^6)

=3x^2{(x^3)^2-(8y^3)^2}. (a^2-b^2=(a+b)(a-b)

=3x^2(x^3-8y^3)(x^3+8y^3)

=>using identity a^3-b^3=(a-b)(a^2+ab+b^2)

and a^3+b^3=(a+b)(a^2-ab+b^2)

=3x^2(x-8y)(x^2+64y^2+8xy)(x+8y)(x^2+64y^2-8xy)

26) (x^4)^3-(y^4)^3

again using the identity we get

=(x^4-y^4)(x^8+y^8+x^4y^4)

=(x^2+y^2)(x^2-y^2)(x^8+y^8+x^4y^4)

=(x+y)(x-y)(x^2+y^2)(x^8+y^8+x^4y^4)

28)(7x)^3-(3y)^3-2(7x-3y)

= (7x-3y)(49x^2+9y^2+21xy)-2(7x-3y)

taking common (7x-3y)

=(7x-3y)(49x^2+9y^2+21xy-2)

Answer 2

Question -

• Factorise -

$\sf{ 1. \ {x}^7 y - 8 x {y} ^ 7 }$

$\sf{ 2. \ {x}^9 - {y}^9 }$

$\sf{ 3. \ 3{x}^{13} - 192 {x}^7 {y}^ 6 }$

$\sf{ 4. \ {x}^{12}- {y}^{12} }$

$\sf{ 5. \: \: 343 {x}^3 - 14x + 6y - 27 {y}^3 }$

Solution -

$\sf{ \bold { Question \ 1 \ - }} \\ \\ \sf{ {x}^7 y - 8 x {y} ^ 7 } \\ \\ \sf{ Taking \ Like \ Terms \ Common \ - } \\ \\ \sf{ Here, \ the \ like \ terms \ are \ xy . } \\ \\ \sf{ Hence \ - } \\ \\ \sf{ => xy( {x}^6 - 8 {y}^6 ) } \: \\ \\ \sf{ = > xy( {( {x}^{2}) }^{3} } \: - \: {( {2y}^{2} )}^{3} ) \: \\ \\ \sf{xy( {x}^{2} - 2 {y}^{2})( {x}^{4} + 2 {x}^{2} {y}^{2} + 4 {y}^{4} )} \\ \\ \sf{ = > xy(x - \sqrt{2} y)( x + \sqrt{2}y)( {x}^{2} + xy + 2 {y}^{2} )( {x}^{2} - xy + 2 {y}^{2} ) }$

$\sf{ \bold { Question \ 2 \ - }} \\ \\ \sf{ \: {x}^{9} - {y}^{9} } \: \\ \\ \sf{ \: = > {( {x}^{3} )}^{3} } - {( {y}^{3} )}^{3} \: \\ \\ \sf{ = > ( {x}^{3} - {y}^{3} )( {x}^{6} + {x}^{3} {y}^{3} + {y}^{6} )} \\ \\ \sf{ = >(x - y)( {x}^{2} + xy + {y}^{2} )({x}^{6} + {x}^{3} {y}^{3} + {y}^{6} ) }$

$\sf{ \bold { Question \ 3 \ - }} \\ \\ \sf{ 3{x}^{13} - 192 {x}^7 {y}^ 6 } \\ \\ \sf{ Taking \ Like \ Terms \ Common \ - } \\ \\ \sf{ Here, \ the \ like \ terms \ are \ 3 {x}^{7} . } \\ \\ \sf{ Hence \ - } \\ \\ \sf{ =>3 {x}^{7} ( {x}^{6} - 64 {y}^{6} ) } \\ \\ \sf{ = > { ( {x}^3 ) }^2 - { ( 8 { y }^3 ) }^2 } \\ \\ \sf{ = >( {x}^{3} - 8 {y}^{3} )( {x}^{3} + 8 {y}^{3} ) } \\ \\ \sf{ = >(x - 2y)( {x}^{2} + 2xy + 4 {y}^{2} )(x +2)( {x}^{2} - 2xy + 4 {y}^{2})}$

$\sf{ \bold{ Question \ 4 \ - } } \\ \\ \sf{ {x}^{12} - {y}^{12} } \\ \\ \sf{ = > {( {x}^{6} )}^{2} - {( {y}^{6} )}^{2} } \\ \\ \sf{ = >( {x}^{6} + {y}^{6})( {x}^{6} - {y}^{6} )} \\ \\ \sf{ = > ( { {(x}^{2} )}^{3} - { {(y}^{2} )}^{3} \: )( { {(x}^{2} )}^{3} + { {(y}^{2} )}^{3} \: ) } \\ \\ \sf{ = > }( {x}^{2} - {y}^{2} )( {x}^{4} + {x}^{2} {y}^{2} + {y}^{4} )( {x}^{2} + {y}^{2} )( {x}^{4} - {x}^{2} {y}^{2} + {y}^{4} ) \\ \\ \sf{(x - y)(x + y) ( {x}^{4} + {x}^{2} {y}^{2} + {y}^{4} )( {x}^{2} + {y}^{2} )( {x}^{4} - {x}^{2} {y}^{2} + {y}^{4} ) }$

$\sf{ \bold{ Question \ 5 \ - } } \\ \\ \sf{343 {x}^3 - 14x + 6y - 27 {y}^3 } \\ \\ \sf{ {7x}^3 - 2(7x - 3y ) - {3y}^3 } \\ \\ \sf{ Hence \ {a}^3 - {b}^3 \ Formula \ - } \\ \\ \sf{ ( 7x - 3y )( 49{x}^2 + 21xy + 9{y}^2 - 2) }$