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Tutorialspoint
71 Views
To do:We have to factorise the given expressions.Solution:We know that, $(a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca$Therefore, (i) $4x^2 + 9y^2 + 16z^2 +12xy-24yz-16xz = (2x)^2 + (3y)^2 + (-4z)^2 + 2 \times (2x) \times (3y) + 2 \times (3y) \times (-4z) + 2 \times (- 4z) \times (2x)$$= (2x + 3y - 4z)^2$Hence $4x^2 + ... Read More
Tutorialspoint
38 Views
To do:We have to write the given cubes in expanded form.Solution:We know that, $(a+b)^3=a^3+b^3+3ab(a+b)$$(a-b)^3=a^3-b^3-3ab(a-b)$Therefore, (i) $(2 x+1)^{3}=(2x)^3 + 1^3 + 3(2x)(1)(2x + 1)$$= 8x^3 + 1 + 6x (2x + 1)$$= 8x^3 + 1 + 12x^2 + 6x$$= 8x^3 + 12x^2 + 6x + 1$Hence $(2 x+1)^{3}=8x^3 + 12x^2 + ... Read More
Tutorialspoint
3K+ Views
To do:We have to evaluate the given expressions using suitable identities.Solution:We know that, $(a+b)^3=a^3 + b^3 + 3ab(a+b)$$(a-b)^3= a^3-b^3-3ab(a-b)$Therefore, (i) $(99)^3 = (100 - 1)^3$$= (100)^3 - (1)^3 - 3 \times 100 \times 1 (100 - 1)$$= 1000000 - 1 - 300 \times 99$$= 1000000 - 1 - 29700$$= 1000000 ... Read More
Tutorialspoint
52 Views
To do:We have to factorise each of the given expressions.Solution:We know that, $(a+b)^3=a^3+b^3+3ab(a+b)$$(a-b)^3=a^3-b^3-3ab(a-b)$Therefore, (i) $8a^3 + b^3 + 12a^2b + 6ab^2=( 2a)^3 + (b)^3 + 3( 2a)( b)( 2a + b)$ $= (2a + b)^3$Hence $8a^3 + b^3 + 12a^2b + 6ab^2=(2a + b)^3$.(ii) $8 a^{3}-b^{3}-12 a^{2} b+6 a b^{2}=( ... Read More
Tutorialspoint
81 Views
To do:We have to verify (i) \( x^{3}+y^{3}=(x+y)\left(x^{2}-x y+y^{2}\right) \)(ii) \( x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right) \)Solution:(i) We know that, $(x+y)^3 = x^3+y^3+3xy(x+y)$This implies, $x^3+y^3 = (x+y)^3-3xy(x+y)$Taking $(x+y)$ common, we get, $x^3+y^3 = (x+y)[(x+y)^2-3xy]$$x^3+y^3 = (x+y)[(x^2+y^2+2xy)-3xy]$ [Since $(x+y)^2=x^2+2xy+y^2$]$x^3+y^3 = (x+y)(x^2+y^2+2xy-3xy)$$x^3+y^3 = (x+y)(x^2-xy+y^2)$Hence verified. (ii) We know that, $(x-y)^3 = x^3-y^3-3xy(x-y)$This ... Read More
Tutorialspoint
27 Views
To do:We have to factorise each of the given expressions.Solution:We know that, $a^3 + b^3=(a+b)^3-3ab(a+b)$$=(a+b)[(a+b)^2-3ab]$$=(a+b)[a^2+b^2+2ab-3ab]$$=(a+b)(a^2+b^2-ab)$............(i)$ a^3-b^3=(a-b)^3+3ab(a-b)$$=(a-b)[(a-b)^2+3ab]$$=(a-b)[a^2+b^2-2ab+3ab]$$=(a-b)(a^2+b^2+ab)$............(ii)Therefore, (i) $27y^3+125z^3$ can be written as, $27y^3+125z^3=(3y)^3+(5z)^3$This implies, $(3y)^3+(5z)^3= (3y+5z)[(3y)^2-(3y)(5z)+(5z)^2]$ [Using (i)]$= (3y+5z)(9y^2-15yz+25z^2)$Hence $27y^3+125z^3=(3y+5z)(9y^2-15yz+25z^2)$.(ii) $64m^3-343n^3$ can be written as, $64m^3-343n^3 = (4m)^3-(7n)^3$$= (4m-7n)[(4m)^2+(4m)(7n)+(7n)^2]$ [Using (ii)]$= (4m-7n)(16m^2+28mn+49n^2)$Hence $64m^3-343n^3= (4m-7n)(16m^2+28mn+49n^2)$.Read More
Tutorialspoint
39 Views
Given:\( 27 x^{3}+y^{3}+z^{3}-9 x y z \)To do:We have to factorize the given expression.Solution:We know that, $a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$Therefore, $27 x^{3}+y^{3}+z^{3}-9 x y z = (3x)^3 + (y)^3 + (z)^3 ... Read More
Tutorialspoint
81 Views
To do:We have to verify that \( x^{3}+y^{3}+z^{3}-3 x y z=\frac{1}{2}(x+y+z)\left[(x-y)^{2}+(y-z)^{2}+(z-x)^{2}\right] \).Solution:We know that, $a^3 + b^3 + c^3 - 3abc = (a + b + c) (a^2 + b^2 + c^2 - ab - bc - ca)$Therefore, LHS $=x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)(x^2 + y^2 + z^2-xy-yz-zx)$$= \frac{1}{2}(x+ y+ z)[2(x^2 ... Read More
Tutorialspoint
51 Views
Given: $x+y+z = 0$To do:We have to show that $x^{3}+y^{3}+z^{3}=3xyz$Solution:We know that, $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$Therefore, $x^{3}+y^{3}+z^{3}-3xyz= (0)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$ [Since $x+y+z =0$]$x^{3}+y^{3}+z^{3}-3xyz=0$$x^{3}+y^{3}+z^{3}=3xyz$ Hence proved.Read More
Tutorialspoint
42 Views
To do:We have to find the values of each of the given expressions.Solution:We know that, $x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$If $x+y+z =0$, then$x^{3}+y^{3}+z^{3}-3xyz= (0)(x^{2}+y^{2}+z^{2}-xy-yz-zx)$$x^{3}+y^{3}+z^{3}-3xyz=0$$x^{3}+y^{3}+z^{3}=3xyz$ (i) \( (-12)^{3}+(7)^{3}+(5)^{3} \)Here, $a=-12, b=7, c=5$This implies, $a+b+c=-12+7+5$$=-12+12$$=0$Therefore, $(-12)^{3}+(7)^{3}+(5)^{3}=3(-12)(7)(5)$$=-36\times35$$=-1260$(ii) \( (28)^{3}+(-15)^{3}+(-13)^{3} \)$a=28, b=-15, c=-13$This implies, $a+b+c=28+(-15)+(-13)$$=28-(15+13)$$=28-28$$=0$Therefore, $(28)^{3}+(-15)^{3}+(-13)^{3}=3(28)(-15)(-13)$$=84\times195$$=16380$Read More