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with a,b∈R prove that ((a+(√3)bi))^(1/3) +((a−(√3)bi))^(1/3) has always real value and find this value (or a way how to find). examples: a=15, b=((28)/9) ⇒ result=5 a=6, b=((35)/9) ⇒ result=4 a=−24, b=((80)/9) ⇒ result=4

$${with}\:{a},{b}\in{R}\:{prove}\:{that} \\ $$$$\sqrt[{\mathrm{3}}]{{a}+\sqrt{\mathrm{3}}{bi}}+\sqrt[{\mathrm{3}}]{{a}−\sqrt{\mathrm{3}}{bi}} \\ $$$${has}\:{always}\:{real}\:{value}\:{and}\:{find}\:{this} \\ $$$${value}\:\left({or}\:{a}\:{way}\:{how}\:{to}\:{find}\right). \\ $$$${examples}: \\ $$$${a}=\mathrm{15},\:{b}=\frac{\mathrm{28}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{5} \\ $$$${a}=\mathrm{6},\:{b}=\frac{\mathrm{35}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{4} \\ $$$${a}=−\mathrm{24},\:{b}=\frac{\mathrm{80}}{\mathrm{9}}\:\:\Rightarrow\:{result}=\mathrm{4} \\ $$

Question Number 78814 Answers: 1 Comments: 1

Question Number 78820 Answers: 1 Comments: 0

please what is the fomula to determinate the equations of bissectors in triangle?

$$\mathrm{please}\:\mathrm{what}\:\mathrm{is}\:\mathrm{the}\:\mathrm{fomula}\:\mathrm{to}\: \\ $$$$\mathrm{determinate}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\: \\ $$$$\mathrm{bissectors}\:\mathrm{in}\:\mathrm{triangle}? \\ $$

Question Number 78762 Answers: 1 Comments: 1

3acr^2 (1−r)+3apr(1−r)(pa+qb) +3bqr(1−r)(pa+qb) = 3(1−r)^2 (pa+qb)^2 +r^2 b^2 Find p, q, r such that the equation is satisfied for general any values of a,b,c.

$$\mathrm{3}{acr}^{\mathrm{2}} \left(\mathrm{1}−{r}\right)+\mathrm{3}{apr}\left(\mathrm{1}−{r}\right)\left({pa}+{qb}\right) \\ $$$$+\mathrm{3}{bqr}\left(\mathrm{1}−{r}\right)\left({pa}+{qb}\right) \\ $$$$\:\:\:=\:\mathrm{3}\left(\mathrm{1}−{r}\right)^{\mathrm{2}} \left({pa}+{qb}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$${Find}\:{p},\:{q},\:{r}\:{such}\:{that}\:{the}\:{equation} \\ $$$${is}\:{satisfied}\:{for}\:{general}\:{any} \\ $$$${values}\:{of}\:{a},{b},{c}.\: \\ $$

Question Number 78755 Answers: 0 Comments: 2

Solve the equation: xy + 5x + 5y = − 25 ... (i) yz + 3y + 5z = − 15 ... (ii) xz + 5z + 3x = − 15 ... (iii)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}: \\ $$$$\:\:\:\:\mathrm{xy}\:+\:\mathrm{5x}\:+\:\mathrm{5y}\:\:=\:\:−\:\mathrm{25}\:\:\:\:\:\:...\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\mathrm{yz}\:+\:\mathrm{3y}\:+\:\mathrm{5z}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:...\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\mathrm{xz}\:+\:\mathrm{5z}\:+\:\mathrm{3x}\:\:=\:\:−\:\mathrm{15}\:\:\:\:\:\:...\:\left(\mathrm{iii}\right) \\ $$

Question Number 78732 Answers: 0 Comments: 2

Question Number 78694 Answers: 6 Comments: 2

Solve the equation. x^2 − (y − z)^2 = 10 ... (i) y^2 − (z − x)^2 = 5 ... (ii) z^2 − (x − y)^2 = 2 ... (iii)

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation}. \\ $$$$\:\:\:\:\:\mathrm{x}^{\mathrm{2}} \:−\:\left(\mathrm{y}\:−\:\mathrm{z}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{10}\:\:\:\:\:\:...\:\left(\mathrm{i}\right) \\ $$$$\:\:\:\:\:\mathrm{y}^{\mathrm{2}} \:−\:\left(\mathrm{z}\:−\:\mathrm{x}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{5}\:\:\:\:\:\:...\:\left(\mathrm{ii}\right) \\ $$$$\:\:\:\:\:\mathrm{z}^{\mathrm{2}} \:−\:\left(\mathrm{x}\:\:−\:\mathrm{y}\right)^{\mathrm{2}} \:\:=\:\:\mathrm{2}\:\:\:\:\:\:...\:\left(\mathrm{iii}\right) \\ $$

Question Number 78650 Answers: 2 Comments: 4

Question Number 78568 Answers: 1 Comments: 0

Find the roots of the equation bx^3 − (3b + 2)x^2 − 2(5b − 3)x + 20 = 0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\mathrm{bx}^{\mathrm{3}} \:−\:\left(\mathrm{3b}\:+\:\mathrm{2}\right)\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{2}\left(\mathrm{5b}\:−\:\mathrm{3}\right)\mathrm{x}\:+\:\mathrm{20}\:\:=\:\:\mathrm{0} \\ $$

Question Number 78563 Answers: 0 Comments: 0

Question Number 78503 Answers: 3 Comments: 0

if x,y >1 prove (x^2 /(y−1))+(y^2 /(x−1))≥8

$${if}\:{x},{y}\:>\mathrm{1}\: \\ $$$${prove}\:\frac{{x}^{\mathrm{2}} }{{y}−\mathrm{1}}+\frac{{y}^{\mathrm{2}} }{{x}−\mathrm{1}}\geqslant\mathrm{8} \\ $$

Question Number 78465 Answers: 1 Comments: 0

Question Number 78460 Answers: 3 Comments: 3

Σ_(n=1) ^∞ (n^3 /3^n )=?

$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\:\frac{{n}^{\mathrm{3}} }{\mathrm{3}^{{n}} }=? \\ $$

Question Number 78449 Answers: 1 Comments: 1

find the domain of definition of f(x)=((−x)/(∣x∣−x))

$$\mathrm{find}\:\mathrm{the}\:\mathrm{domain}\:\mathrm{of}\:\mathrm{definition}\:\mathrm{of}\: \\ $$$$\mathrm{f}\left({x}\right)=\frac{−{x}}{\mid{x}\mid−{x}} \\ $$

Question Number 78444 Answers: 0 Comments: 0

Question Number 78440 Answers: 0 Comments: 0

ab = p (a−2b)(a+b)(a+2b)=−24q Find ((a+b)/3) in terms of p, q.

$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{ab}\:=\:{p} \\ $$$$\:\:\left({a}−\mathrm{2}{b}\right)\left({a}+{b}\right)\left({a}+\mathrm{2}{b}\right)=−\mathrm{24}{q} \\ $$$${Find}\:\:\:\frac{{a}+{b}}{\mathrm{3}}\:\:\:\:{in}\:{terms}\:{of}\:{p},\:{q}. \\ $$

Question Number 78399 Answers: 2 Comments: 14

dear sir W, Mjs the set {1,4,n} have the condition that if two different elements are selected and 2112 is added to the result , then the result is a perfect square if n is a positif number . then the number of possible values of n is (A) 8 (B) 7 (C) 6 (D) 5 (E) 4

$${dear}\:{sir}\:{W},\:{Mjs}\: \\ $$$${the}\:{set}\:\left\{\mathrm{1},\mathrm{4},{n}\right\}\:{have}\:{the}\:{condition}\:{that}\: \\ $$$${if}\:{two}\:{different}\:{elements}\:{are} \\ $$$${selected}\:{and}\:\mathrm{2112}\:{is}\:{added}\:{to} \\ $$$${the}\:{result}\:,\:{then}\:{the}\:{result}\: \\ $$$${is}\:{a}\:{perfect}\:{square}\:{if}\:{n}\:{is}\:{a}\: \\ $$$${positif}\:{number}\:.\:{then}\:{the}\:{number}\: \\ $$$${of}\:{possible}\:{values}\:{of}\:{n}\:{is}\: \\ $$$$\left({A}\right)\:\mathrm{8}\:\:\:\:\left({B}\right)\:\mathrm{7}\:\:\:\:\:\left({C}\right)\:\mathrm{6}\:\:\:\:\:\left({D}\right)\:\mathrm{5} \\ $$$$\left({E}\right)\:\mathrm{4} \\ $$

Question Number 78390 Answers: 1 Comments: 0

solve for different digits a,b,c,d such that abcd=(ab+cd)^2 .

$${solve}\:{for}\:\boldsymbol{{different}}\:{digits}\:{a},{b},{c},{d}\: \\ $$$${such}\:{that}\:\boldsymbol{{abcd}}=\left(\boldsymbol{{ab}}+\boldsymbol{{cd}}\right)^{\mathrm{2}} . \\ $$

Question Number 78358 Answers: 0 Comments: 1

Q. solve (2^(sin^2 (x)) /(sin^2 (x) )) + (3^(cos^2 (x)) /(cos^2 (x))) = 6

$${Q}.\:{solve} \\ $$$$ \\ $$$$\frac{\mathrm{2}^{{sin}^{\mathrm{2}} \left({x}\right)} }{{sin}^{\mathrm{2}} \left({x}\right)\:}\:+\:\frac{\mathrm{3}^{{cos}^{\mathrm{2}} \left({x}\right)} }{{cos}^{\mathrm{2}} \left({x}\right)}\:=\:\mathrm{6} \\ $$

Question Number 78357 Answers: 2 Comments: 7

Show that: (((√(1 + 6x)) − (1/(√(1 − 6x))))/((√(1 + 3x)) − (1/(√(1 − 3x))))) = 4 + 6x Ignoring higher power of x in the expansion

$$\mathrm{Show}\:\mathrm{that}:\:\:\:\frac{\sqrt{\mathrm{1}\:+\:\mathrm{6x}}\:\:−\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:\mathrm{6x}}}}{\sqrt{\mathrm{1}\:+\:\mathrm{3x}}\:\:−\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}\:−\:\mathrm{3x}}}}\:\:\:\:\:=\:\:\:\mathrm{4}\:+\:\mathrm{6x} \\ $$$$\mathrm{Ignoring}\:\mathrm{higher}\:\mathrm{power}\:\mathrm{of}\:\:\mathrm{x}\:\:\mathrm{in}\:\mathrm{the}\:\mathrm{expansion} \\ $$

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