Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y

Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

Question

Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

Aman Bansal · Answer

Dear Abhinav,

we can write x+(1/z) as (1/yz) + (1/z) solving this we get(1+y)/yz = 5 ................(1)

now from the second equation we get y + yz = 29

or, yz = 29-y putting it in (1) we get (1+y) = 5(29-y)

or, 1+y = 145 - 5y6y = 144or, y = 24now again putting it in (1) we get 25 = 120z

or, z = 5/24so x = 1/yz or, x = 1/5 notice that this fulfills the requirement xyz = 1so now z + 1/x = z + zy = 5/24 + (5/24)(1/5) = 5/24 + 1/24 = 1/4 answer.

BEST OF LUCK..!!!!

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Rishabh Raj · Answer

xyz=1, thus yz=1/x. x+1/z=5 so z=1/[5-x ] y+1/x=29 so y=[ 29x-1 ]/x Now, yz=1/x [ [29x-1 ]/x]*[1/[5-x ]] =1/x On solving, x=1/5 yz=5, y=24 z+1/y=[yz+1]/y. Substitute the values of yz and y. The answer u wud get as 1/4.

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Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y

Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

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Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.

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Q.1 If xyz=1, x + 1/z =5, y + 1/x =29, then find z + 1/y

Q.2 If a+b+c=2, a^2 + b^2 + c^2 =6, a^3 + b^3 + c^3 =8, then find a^4 + b^4 + c^4.