Algebra - Study Mode
[#191] If $$
oot 3 of a +
oot 3 of b =
oot 3 of c { ext{,}}$$ xa0xa0 then the simplest value of $${left( {a + b - c}
ight)^3}$$ xa0 + $$27abc$$ xa0 is?
Correct Answer
(D) 0
Explanation
Solution: $$eqalign{
&
oot 3 of a +
oot 3 of b =
oot 3 of c cr
& { ext{Take cube both sides}} cr
& {left( {
oot 3 of a +
oot 3 of b }
ight)^3} = {left( {
oot 3 of c }
ight)^3} cr
& Rightarrow a + b + 3{a^{frac{1}{3}}}{b^{frac{1}{3}}}left( {
oot 3 of a +
oot 3 of b }
ight) = c cr
& Rightarrow a + b + 3{a^{frac{1}{3}}}{b^{frac{1}{3}}}{c^{frac{1}{3}}} = c cr
& Rightarrow a + b - c = - 3{a^{frac{1}{3}}}{b^{frac{1}{3}}}{c^{frac{1}{3}}} cr
& { ext{Again take cube both sides}} cr
& Rightarrow {left( {a + b - c}
ight)^3} = - 27abc cr
& Rightarrow {left( {a + b - c}
ight)^3} + 27abc = 0 cr
& cr
& {x08f{Alternate:}} cr
& { ext{Put value of }} cr
& a = 0 cr
& b = 1 cr
& c = 1 cr
& { ext{Value of }}{left( {a + b - c}
ight)^3} + 27abc cr
& = {left( {0 + 1 - 1}
ight)^3} + 27 imes 0 imes 1 imes 1 cr
& = 0 cr} $$
[#192] If $$x =
oot 3 of {a + sqrt {{a^2} + {b^3}} } $$ xa0 xa0 + $$
oot 3 of {a - sqrt {{a^2} + {b^3}} } $$ xa0 xa0 then $${x^3} + 3bx$$ xa0 is equal to?
Correct Answer
(C) 2a
Explanation
Solution: $${ ext{Given}},x =
oot 3 of {a + sqrt {{a^2} + {b^3}} } ,,+ $$ xa0 xa0 xa0$$
oot 3 of {a - sqrt {{a^2} + {b^3}} } $$ $$eqalign{
& { ext{Let}},z = sqrt {{a^2} + {b^3}} cr
& herefore x =
oot 3 of {a + z} +
oot 3 of {a - z} cr
& { ext{Cubing both side}} cr} $$ $$ herefore {x^3} = {left( {
oot 3 of {a + z} }
ight)^3} + {left( {
oot 3 of {a - z} }
ight)^3} ,
+ $$ xa0 xa0 xa0$$,3{left( {a + z}
ight)^{frac{1}{3}}}$$ xa0$${left( {a - z}
ight)^{frac{1}{3}}} imes, $$ xa0 $$left( {
oot 3 of {a + z} +
oot 3 of {a - z} }
ight)$$ $$ ,Rightarrow {x^3} = a + z + a - z ,+, $$ xa0xa0 $$3{left( {{a^2} + az - az - {z^2}}
ight)^{frac{1}{3}}}$$ xa0xa0 $$ imes left( x
ight)$$ $$eqalign{
& Rightarrow {x^3} = 2a + 3{left( {{a^2} - {z^2}}
ight)^{frac{1}{3}}} imes left( x
ight) cr
& { ext{Put the value of }}z cr
& Rightarrow {x^3} = 2a + 3{left( {{a^2} - {a^2} - {b^3}}
ight)^{frac{1}{3}}} imes left( x
ight) cr
& Rightarrow {x^3} = 2a + 3{left( { - {b^3}}
ight)^{frac{1}{3}}} imes left( x
ight) cr
& Rightarrow {x^3} = 2a - 3bx cr
& herefore {x^3} + 3bx = 2a cr} $$
[#193] If $$frac{{{x^{24}} + 1}}{{{x^{12}}}} = 7,$$ xa0xa0 then the value of $$frac{{{x^{72}} + 1}}{{{x^{36}}}} = ,?$$
Correct Answer
(D) 322
Explanation
Solution: $$eqalign{
& frac{{{x^{24}} + 1}}{{{x^{12}}}} = 7{ ext{ }}left( {{ ext{Given}}}
ight) cr
& Rightarrow frac{{{x^{24}}}}{{{x^{12}}}} + frac{1}{{{x^{12}}}} = 7 cr
& Rightarrow {x^{12}} + frac{1}{{{x^{12}}}} = 7 cr
& { ext{Cubing both sides}} cr
& Rightarrow {left( {{x^{12}} + frac{1}{{{x^{12}}}}}
ight)^3} = {7^3} cr} $$ $$ Rightarrow {x^{36}} + frac{1}{{{x^{36}}}} + frac{{3 imes {x^{12}} imes 1}}{{{x^{12}}}}$$ xa0 xa0 $$left( {{x^{12}} + frac{1}{{{x^{12}}}}}
ight) = $$ xa0xa0 $$343$$ $$eqalign{
& Rightarrow {x^{36}} + frac{1}{{{x^{36}}}} + 3left( 7
ight) = 343 cr
& Rightarrow {x^{36}} + frac{1}{{{x^{36}}}} = 343 - 21 cr
& Rightarrow {x^{36}} + frac{1}{{{x^{36}}}} = 322 cr
& Rightarrow frac{{{x^{72}} + 1}}{{{x^{36}}}} = 322 cr} $$
[#194] If x = 2 then the value of x 3 + 27x 2 + 243x + 631 = ?
Correct Answer
(B) 1233
Explanation
Solution: $$eqalign{
& { ext{Given, }}x = 2 cr
& { ext{Find }}{x^3} + 27{x^2} + 243x + 631 cr
& { ext{To put value }}x = 2{ ext{ }} cr
& Rightarrow {2^3} + 27 imes {2^2} + left( {243 imes 2}
ight) + 631 cr
& Rightarrow 8 + 108 + 486 + 631 cr
& Rightarrow 1233 cr} $$
[#195] If 5x + 9y = 5 and 125x 3 + 729y 3 = 120, then the value of the product of x and y is?
Correct Answer
(C) $$frac{1}{{135}}$$
Explanation
Solution: 5x + 9y = 5 . . . . . (i) 125x 3 + 729y 3 = 120 . . . . . (ii) From equation (i) cubing both sides ⇒ (5x + 9y) 3 = 5 3 ⇒ 125x 3 + 729y 3 + 3 × 5x × 9y(5x + 9y) = 125 ⇒ 125x 3 + 729y 3 + 135xy × 5 = 125 ⇒ 120 + 135xy × 5 = 125 ⇒ 135xy × 5 = 5 ⇒ xy = $$frac{1}{{135}}$$ ∴ product of x & y = $$frac{1}{{135}}$$