Let, x be the number. Then,           ∛x = 3 Cubing both sides,             x = 33 ∴ x = 27. Therefore, required no.is .....

Let, x be the number . Then,     √x = 4 Squaring both sides,      x = 42 or, x = 16. Therefore 16 is required .....

Solution: Let, x and y be the present ages of mother and daughter respectively. Then,         x6  = y   or, x = 6y  ………….1). And, x - y = .....

Solution: We have,                     a3/(a-b) - b3/(a-b)                 = (a3 - b3) / (a - b)       .....

Solution: We have,     a4 / (a-b) - b4 / (a-b) = (a4-b4) / (a-b) = ((a-b) (a+b) (a2 + b2)) / (a-b) = (a+b)(a2+b2), which is required .....

Solution: We have,      (2x+6) / (x2-9) ÷ (3x2+9x)/(2x2-6x)   = (2x+6) / (x2-9) × (2x2-6x) / (3x2+9x)   = [(2(x+3)) / ((x+3)(x-3))] × [(2x(x-3)) / (3x(x+3))]   = 4 / (3(x+3)), which is .....

Solution: We have,      x/(y(x-y))-y/(x(x-y))            = (x2-y2)/(xy(x-y))    = ((x-y)(x+y))/(xy(x-y))         = .....

Solution: We have,    (x2-xy) / (x2+xy) ÷ (x2 (x-y)) / (x3+x2 y) = (x2-xy) / (x2 + xy)  × (x3+x2 y) / (x2 (x-y)) = (x(x-y)) / (x(x + y)) × (x2 (x+y)) / (x2 (x-y)) = 1 .....

Solution:  For the given expression to be undefined, its denominator should be 0. That is,      5A-5=0 or, 5A=5 ∴     A=1,  which is required .....

Solution:    a-bab+b-cbc+c-aca = ca-b+ab-c+bc-aabc = ac-bc+ab-ac+bc-ababc =0  , which is required .....

1-1P1-1P-1(1+2P-2 ) = P-1PP-1-1P-1(P-2+2P-2 ) = P-1PP-2P-1(PP-2 ) = 1, which is required .....

Solution: a-ba+b-a+ba-b+2aba2-b2 = a-ba+b-a+ba-b+2aba-ba+b = a-b2-a+b2+2aba-ba+b = a-b-a-ba-b+a+b+2aba2-b2 = -2b.2a+2aba2-b2 = .....

Solution: x2yx-y+y2xy-x = x2yx-y-y2xx-y = x3-y3xyx-y = x-yx2+xy+y2xyx-y = x-yx2+xy+y2xyx-y = ( x2 + xy + y2 ) / xy , which is required .....

Solution: =   x2-xyx2+xy÷x2x-yx3+x2y = xx-yxx+y÷x2x-yx2x+y  =1 , which is required .....

Solution: For the given expression to be undefined, its denominator should be 0. That is,     x-5=0 ∴ x=5, which is required .....

81m29m-7n+49n27n-9m = 9m29m-7n-7n29m-7n = 9m2-7n29m-7n = 9m-7n9m+7n9m-7n = 9m+7n  , which is required .....

Solution: x2-xy+y2x-y-x2+xy+y2x+y = (x+y)(x2-xy+y2)-x-y(x2+xy+y2)x-yx+y = x3+y3-x3-y3x-yx+y = 2y3     x2-y2    .....

Solution:  aa2-2ab+b2+ba2-b2 = aa-b2+ba-ba+b = aa+b+ba-ba-b2a+b = a2+ab+ab-b2a-b2a+b = a-b2a-b2a+b = .....

= 1+ y+ 1-y41- y(1- y + 2y41- y = 2 41- y(1- y + 2y41- y1+ y    [ Since,1 - y = 12 - √y2 = ( 1 -√ y) (1 + √ y).] = 2+2y41- y1- y = 21 +y41- y1- y = 121- .....

= x+3 x2-3x+9+ x-3x2+ 3x+9(x2+ 3x+9)(x2-3x+9) - 54x2+92+ 9x2 = (x3+ 27)+(x3-27)(x2+ 3x+9)(x2-3x+9) - 54x2+92-18x2+9x2 = 2x3(x2+ 3x+9)(x2-3x+9) - 54x2+92-9x2 = 2x3(x2+ 3x+9)(x2-3x+9) - 54x2+3x+9x2- 3x+9 = 2x3- 54(x2+ 3x+9)(x2-3x+9) = 2(x3-27)(x2+ .....

= 2x-2x-3 - 2x-1x-3 + 1x-1x-2 = 2x-1- 2x-2+ x-3x-1x-2x-3 = 2x-2- 2x+4+ x-3x-1x-2x-3 = x-1x-1x-2x-3 = .....

Solution: We .....

Solution: We have,  1a2-5a+6-2a2-4a+3-1a2-3a+2 =1a2-2a-3a+6-2a2-3a-a+3-1a2-2a-a+2 .....

= 2a-3a-4a-5 - a-1a-3a-4- a-2a-5a-3 = 2a-3a-3- a-1a-5- a-2a-4a-3a-4a-5 = 2a2- 6a+9- a2- 6a+5-a2- 6a+8a-3a-4a-5 = 2a2- 12a+18- a2- 6a+5-a2- 6a+8a-3a-4a-5 = - 12a+18- - 6a+5-- 6a+8a-3a-4a-5 = .....

Solution: We have,           ((a-b)/ab) + ((b-c)/bc) + ((c-a)/ca)       =  (c(a-b) + a(b-c) + b(c-a))/abc         =  (ca-cb + ab-ca + .....

Solution: We have,        a2/(a-b) + b2 / (b-a) =  a2 / (a-b) -  b2 / (a-b)       = (a2-b2) / (a-b) = (a-b)(a+b)/ (a-b)     = (a+b),which is .....

4solution: = 2x+6x2-9+3x2+9x2x2-6x .....

= a-b-ca-b+ca+b+ca-b-c + b-c-ab-c+ab+c+ab-c-a + c-a-bc-a+bc+b+cc-a-b = a-b+ca+b+c + b-c+aa+b+c + c-a+ba+b+c = a+c+ba+b+c = .....

Here,    (xa/x-b)a-b ×  (xb/x-c )b-c ×  (xc/x-a)c+a = x(a+b)(a-b)× x(b+c)(b-c) × x(c+a)(c-a) = xa2-b2 × xb2-c2 ×xc2-a2 = xa2-b2+b2-c2+c2-a2 = x0 = 1 . .....

Here,    (xa/xb )a2+b2+ab × (xb/xc ) b2+bc+c2 × (xc/xa )c2+ca+a2 = x(a-b)(a2+ab+b2) ×x(b-c)(b2+bc+c2 ) × x(a-b)(c2 +ca+a2) = xa3-b3 × xb3-c3× xc3-a3 = xa3-b3+ b3-c3+ c3-a3 = x0 = 1 .....

Given,          p = ax, q = ay, r = az To prove: py-zqz-xrx-y = 1 LHS = py-zqz-xrx-y         = (ax) y-z(ay)z-x(az)x-y         = axy-xz . ayx-xy. axz-yz       .....

Here, Given,          x = 31/3 - 32/3 To prove:          x3 – 9x + 6 = 0. LHS = x3 – 9x + 6          = (31/3 - 32/3)3 – 9(31/3 - 32/3) + 6     .....

Here, Given,          x = 21/3 + 2-1/3 To prove: 2x3 – 6x = 5. LHS = 2x3 – 6x          = 2 (21/3 + 2-1/3)3- 6 (21/3 + 2-1/3)          = 2 {(21/3)3 + .....

Here, Given,           a + b + c = 0. To prove:             x1/bc. x1/ca. x1/ab =1 LHS = x1/bc. x1/ca. x1/ab           = x1/bc + 1/ca+1/ab     .....

LHS =[(1/(x )+y)a  ( 1/x- y)a] / [(1/y+ x)a ( 1/y- x)a]         = [{(1+xy)/(x )}a  ( (1-xy)/x)a] / [{1+xy)/y)a  ( (1-xy)}/y)a ]         = [{(1+xy)/x}/{(1+xy)/y}]a ⨉ .....

Here,             21-x = ax-1      Or,  21-x / ax-1 = 1       Or,  21-x . a1-x = 1      Or,  (2a)1-x = 1      Or,  (2a)1-x = .....

Here,   2x+2 + (2(x+3)/2) = 1 Or, 2x+2 + 2x+3-1 = 1 Or, 2x+2 + 2x+2 = 1 Or, 2. 2x+2 = 1 Or, 2x+2+1 = 1 Or, 2x+3 = 1 Or, 2x+3 = 20 Or,   x+3 = 0 Or,x = -3    ∴ x = -3 .....

Here, 3x3 – 9x = 3(31/3 + 3-1/3)3 – 9(31/3 + 3-1/3)                = 3 ( 3 + 3 . 32/3. 3-1/3 + 3 . 31/3. 3-2/3 + 3-1 ) – 9(31/3 + 3-1/3)               .....

Here, LHS = (x(l-n)/x(l-m) )(m+n) ⨉ (x(m-l)/x(m-n) )(n+l) ⨉ (x(n-m)/x(n-l) )(l+m)            = (x l-n-l+m) m+n (x m-l-m+n)n+l (xn-m-n+l)l+m            = (xm-n)m+n (xn-l)n+l .....

Here, LHS = (x(1/(a-b )))(a/(c-a)) ⨉ (x(1/(b-c)))(b/(a-b)) ⨉ (x(1/(c-a)))(c/(b-c)) = x(a/((a-b)(c-a) )) ⨉ x(b/((b-c)(a-b) )) ⨉ x(c/((c-a)(b-c) ))  = x(a/((a-b)(c-a) )+ b/((b-c)(a-b) )+c/((c-a)(b-c) ))   = .....

Here,    4x.2x+1-2x / (2(2x-1).2(x+2) - 2x ) = 22x.2x.21-2x / (22x.2-1.2x.22-2x ) = 2x (22x.21-1) / {2x (22x.2-1+2-1)} = (22x.21-1) / (22x.21-1) = 1 .....

Here, LHS = (xp/xq )(p+q) ⨉ (xq/xr )(q+r) ⨉ (xr/xp )(r+p)          = (xp-q)p+q . (xq-r)q+r.(xr-p)r+p          = x p2-q2.X q2-r2.X r2-p2         .....

Here, LHS = (x(a+b)/xc )(a-b) ⨉ (x(b+c)/xa )(b-c) ⨉ (x(c+a)/xb )(c-a)           = (xa+b-c)a-b (xb+c-a)b-c (xc+a-b)c-a           = x (a-b)(a+b-c) x(b-c)(b+c-a) x(c-a)(c+a-b)     .....

Here,       2.a3x-1 = 8x Or, 2.a3x-1 = (23)x Or, 2.a3x-1 = 23x Or, a3x-1 = 23x-1 Or, a(3x-1)/2(3x-1)  = 1 Or, (a/2)(3x-1) = 1 Or,  (a/2)(3x-1) = (a/2)0 Equating the exponents,       3x -1 = 0 Or, 3x  = .....

LHS = 1/[1+ x(l-m) +   x(n-m) ] + 1/[1+ x(m-n) +   x(l-n) ]+ 1/[1+ x(n-l) +  x(m-l) ]         = 1/[1+ xl/(xm )+xn/(xm)] + 1/[1+ xm/(xn )+xl/xn] + 1/[1+ xn/(xl)+xm/xl]      .....

Here,            32x+1 = 9x+2 – 26    Or, 32x+1 = 32(x+2) – 26    Or, 32x+1 = 32x+4 – 26    Or, 32x+1 - 32x+4 = -26    Or, 32x.31 – 32x.34 = -26   .....

Here, LHS = (xa/xb )c  ⨉ (xb/xc )a  ⨉ (xc/xa )b          = (xa-b)c  (xb-c)a (xc-a)b          = xac – bc xab – ac xbc – ab          = .....

Here, LHS = x(l+m)/x2n   ⨉ x(m+n)/x2l   ⨉ x(n+l)/x2m          = x(l+m-2n) x(m+n-2l) x(n+l-2m)         = x(l+m-2n+m+n-2l+n+l-2m)         = .....

Here, Given,        ax = by = cz and b2 = ac. To prove: 1/x +  1/z = 2/y. Here,        ax = by                 And, by = cz Or,  a = .....

Here,        22x+1 – 5.2x +2 = 0 Or, 22x. 2 – 5.2x + 2 = 0 Or, 2. (2x)2 – 5. 2x + 2 = 0 . Let ,      2x = a Then,        2a2 – 5a + 2 = 0 Or, 2a2 – (4+1)a + 2 = .....