Sum of Algebra

Sum of Algebra 

The collection of Sum of Algebra of different competitive exams is given here.

1) If  {a}^x =  b, {b}^x =c  and  {c}^x = a then find the value of xyz.

{ a }^{ xyz }\quad =\quad { ({ a }^{ x }) }^{ yz }
{ a }^{ xyz }\quad =\quad {b }^{ yz }
{ a }^{ xyz }\quad =\quad {{b}^y}^z
{ a }^{ xyz }\quad =\quad { a }^z
{ a }^{ xyz }\quad = a
on comparison we get
xyz = 1
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2) If a+b+c = 9 and  ab+bc+ca = 40 then find the value of {a}^2+{b}^2+{c}^2.

By formula
{(a+b+c)}^2 = {a}^2 + {b}^2+{c}^2 + 2(ab+bc+ca)
{9}^2 = {a}^2 + {b}^2+{c}^2 + 2\times 40
81 = {a}^2 + {b}^2+{c}^2 + 80
{a}^2 + {b}^2+{c}^2 =81 – 80
{a}^2 + {b}^2+{c}^2 = 1
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3) If a+b=7 then find the value of {(a-b)}^2 +4ab+ 1 .

{(a-b)}^2 +4ab+ 1 .= {a}^2 +{b}^2 – 2ab +4ab+ 1 .
= {a}^2 +{b}^2 + 2ab + 1 .
= {(a+b)}^2 + 1 .
= 49 + 1
= 50.
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4) If 2x+y = 3  and xy = 1 then find the value of {(x+y)}^{x-y}

{(2x-y)}^2 = {(2x+y)}^2 – 4\times 2x \times y
= {3}^2 – 4\times 2\times1
=9 – 8
= 1
2x – y = 1 ————(1)
2x + y = 3 ———–(2)
solving 1 and 2 we get x = 1 and y = 1
Now, {(x+y)}^{x-y}
= {2}^0
=1 {Something to the power Zero is 1}
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5) If   x = (b-c) (a-d),  y = (c-a) (b-d) and z = (a-b) (c-d) then find the value of {x}^3 + {y}^3 +{z}^3 .

x+y+z = (b-c)(a-d)+ (c-a)(b-d) + (a-b)(c-d)
x+y+z = 0
cubing both side we get
{x}^3 + {y}^3 + {z}^3 = 3xyz
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6) If {(a-2)}^2 + {(b+3)}^2 + {(c-4)}^2 = 0 then find the value of   a-b+c

By given
a = 2
b = -3
c = 4
a-b+c = 2-(-3)+4
=2+3+4
9.
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7) If {(x)}^2+{(y)}^2+4x+4y+8 =0 then find the value of x+y .

{(x)}^2+{(y)}^2+4x+4y+8 =0
({x}^2+4x+4)+({y}^2+4y+4) = 0
{(x+2)}^2 + {(y+2)}^2 = 0
x+2 = 0 and y+2 = 0
so x=-2 and y = -2,
therefore x+y = (-2) + (-2)
= -4
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8) If a+b+c = 0,then find the value of {a}^2(b+c) + {b}^2 (c+a) + {c}^2 (a+b) .

{a}^2(b+c) + {b}^2 (c+a) + {c}^2 (a+b) .
= {a}^2(a+b+c-a) + {b}^2 (c+a+b-b) + {c}^2 (a+b+c-c) .
= – ({a}^3+{b}^3+{c}^3)
= – 3abc {a+b+c = 0, then {a}^3+{b}^3+{c}^3 = 3abc
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9) If a+b+c = 13, and ab+bc+ca = 50 then find the value {a}^2 +{b}^2 +{c}^2.

By formula,
{a}^2 +{b}^2 +{c}^2.
= {(a+b+c)}^2 – 2(ab+bc+ca).
= {13}^2 – 2 × 50
= 169 – 100
= 69
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10) If p+q = r and pqr = 30 then find the value of {p}^3 + {q}^3 – {r}^3.

Given that ,
P+q = r
p + q – r = 0
{p}^3 +{q}^3 +{(-r)}^3 = 3pq(-r)
= 3 × (-30)
=-90
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11) If {x}^2+8{y}^2+9{z}^2 – 4xy – 12xz = 0, then which option is correct.
a) 3x = 2y = z
b) x=y=z
c) x=2y=z
d) x+2y+3z=0

{x}^2+8{y}^2+9{z}^2 – 4xy – 12xz = 0
{x}^2 – 4xy+4{y}^2+4{y}^2 – 12xz+9{z}^2 = 0
{(x-2y)}^2 + {(2y-3z)}^2 = 0
Now, x-2y = 0 and 2y-3z = 0.
so, x= 2y = 3z.
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12) If {3}^{(x-y)} = 27, and {3}^{(x+y)} = 243, then find the value of x.

{3}^{(x-y)} = 27
{3}^{(x-y)} = {3}^3
therefore x- y = 3 ——————— (1)
{3}^{(x+y)} = 243
{3}^{(x+y)} = {3}^5
therefore x+y = 5 ————————(2)
on addition of equation 1 and 2 we get
2x = 8
so, x = 4
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13) If x = 12 and y = 4 then find the value of {(x+y)} ^{x/y} .

{(x+y)} ^{x/y}
={(12+4)}^ 3
={16}^3
= 4096.
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14) If {(a-1)}^2 + {(b+2)}^2 + {(c+1)}^2 = 0, then find the value of 2a+3b+7c.

{(a-1)}^2 + {(b+2)}^2 + {(c+1)}^2 = 0
therefore, a-1 = 0 , b+2 = 0, c+1 = 0.
so, a = 1, b = -2, c = -1.
Now, 2a -3b +7c = 2+6-7
= 1
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15) If p = 124 then \sqrt [ 3 ]{ p({ p }^{ 2 }+3p+3)+1 } = ?.

\sqrt [ 3 ]{ p({ p }^{ 2 }+3p+3)+1 }
= {[{(p+1)}^3]}^{1/3}
= (p+1)
=(124+1)
=125
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16) {x}^a.{x}^b.{x}^c = 1 then find the value of {a}^3 +{b}^3+{c}^3.


{x}^a.{x}^b.{x}^c = 1
{x}^{(a+b+c)} = {x}^0
so, a+b+c = 0
Now, {a}^3 +{b}^3+{c}^3 – 3abc = (a+b+c)({a}^2+{b}^2)+{c}^2-ab-bc-ca)
{a}^3 +{b}^3+{c}^3 – 3abc = 0
{a}^3 +{b}^3+{c}^3 = 3abc.
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17) If (x+y+z) = 16 and (xy+yz+zx) = 78 then find the value of {x}^3+{y}^3+{z}^3 -3 xyz.

(x+y+z) = 16
On squaring both side
{(x+y+z)}^2 =256
{x}^2+{y}^2+{z}^2 + 2(xy+yz+zx) = 256
{x}^2+{y}^2+{z}^2 =100
we know that,
{x}^3+{y}^3+{z}^3 -3 xyz = (x+y+z)({x}^2+{y}^2+{z}^2 -xy-yz-zx)
= 16 × (100-78)
= 16 × 22
= 352.
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18) If {a}^2 + {b}^2 + 4{c}^2 = 2(a+b-2c) – 3 and a,b,c are real numbers then find the value of ({a}^2+{b}^2+{c}^2).

{a}^2 + {b}^2 + 4{c}^2 = 2(a+b-2c) – 3
{a}^2 + {b}^2 + 4{c}^2 – 2(a+b-2c) + 3 = 0
{a}^2-2a+1+{b}^2-2b+1+{(2c)}^2 +4c+1 = 0
{(a-1)}^2 +{(b-1)}^2 +{(2c+1)}^2 =0
therefore, {(a-1)}^2 = 0, {(b-1)}^2 =0, {(2c+1)}^2 = 0
so, a=1,b=1, c= 1/2,
Now,
({a}^2+{b}^2+{c}^2) = 1+1+1/4
= 2\frac { 1 }{ 4 }
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19) If {a}^2 +1 =a, then find the value of {a}^{12} + {a}^6+1.

{a}^2 +1 =a
{a}^2 +1 -a=0
Multiplying both side by (a+1),
(a+1)({a}^2 +1 -a)=0
{a}^3+1 = 0
so, a = -1.
Now,
{a}^{12} + {a}^6+1 = {{a}^3}^4 + {{a}^3}^2 + 1
= {(-1)}^4 + {(-1)}2 +1
= 1+1+1
=3
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20) If x = a-b, y = b-c, z = c-a, then find the value of {x}^3+{y}^3 + {z}^3 -3xyz .

x+y+z = (a-b)+(b-c)+(c-a)
x+y+z=0
cubing both side we get,
{x}^3+{y}^3+{z}^3-3xyz = 0
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Sum of Algebra
21) If x=y=333 and z = 334 then find the value of {x}^3+{y}^3+(z)^3-3xyz.

{x}^3+{y}^3+(z)^3-3xyz.
= \frac { 1 }{ 2 }(x+y+z)[{(x-y)}^2 + {(y-z)}^2 + {(z-x)}^2]
=\frac { 1 }{ 2 } <strong><em>×</em></strong> 1000[0+1+1]
= 1000
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22) {a}^2 + {b}^2 + {c}^2 = ab +bc + ca, then find the value of \frac{a+c}{b}

{a}^2 + {b}^2 + {c}^2 = ab +bc + ca
multiplying both sides by 2 we get,
2 {a}^2 +2 {b}^2 + 2{c}^2 = 2ab +2bc +2 ca
({a}^2 – 2ab+{b}^2) + ({b}^2 – 2bc+{c}^2)+({c}^2 – 2ca+{c}^2) = 0
So, {(a-b)}^2 +{(b-c)}^2 + {(c-a)}^2 = 0
Therefore a = b = c
Now, \frac{(a+c)}{b} = 2

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23) If a+b+c = 4 then find the value of {a}^3 + {b}^3 + {c}^3 – 12{c}^2 +48c-64.

a+b+c = 4
a+b+c-4 = 0
We Know that If a+b+c= 0 then, {a}^3 + {b}^3 + {c}^3 = 3abc
So, {a}^3 + {b}^3 + {(c-4)}^3 = 3ab(c-4)
{a}^3 + {b}^3 +[ {c}^3-64-12c(c-4)] = 3abc-12ab
{a}^3 + {b}^3 + {c}^3-64-12{c}^2+48c = 3abc-12ab
{a}^3 + {b}^3 + {c}^3-12{c}^2+48c-64 = 3abc-12ab
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24) 4x -5z = 16 and xz = 12 then find the value of 64{x}^3 – 125{z}^3.

4x -5z = 16
On squaring both side we get,
16{x}^2+25{z}^2 – 40 zx = 256
16{x}^2+25{z}^2 = 256 + 480
16{x}^2+25{z}^2 = 736
16{x}^3-25{z}^3 ={(4x)}^3 – {(5z)}^3
= (4x – 5z)(16{x}^2 + 25{z}^2 + 20xz)
= 16(736+240)
= 15616.
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