Question

if a minus b is equal to 7 and a square + b square is equal to 85 then find the value of a cube minus b cube

Answer 1

$a^3-b^3=721$

Solution:

As given, a minus b is equal to 7

$(a-b) =7$ -------- (i)

Again, a square + b square is equal to 85

$a^2+b^2=85$ -------------- (ii)

Now if we square equation (i), we get,

$(a-b)^2= 7^2$

$a^2-2ab+b^2=49$

$-2ab + 85 = 49$

$-2ab = -36$

ab =18

Now,

$a^3-b^3= (a-b)(a^2+ab +b^2)$

$= 7\times(85+18)$=721

Answer 2

Answer:

$Value\:of \: a^{3}-b^{3} = 721$

Step-by-step explanation:

$Given \: a-b = 7 \: ---(1)\\and \\a^{2}+b^{2}=85\:---(2)$

$We \: know \: the \: algebraic \: identity:\\</p><p>a^{2}+b^{2}-2ab = (a-b)^{2}$

$\implies 85 - 2ab = 7^{2}$

/* From (1) and (2) */

$\implies -2ab = 49 - 85$

$\implies -2ab = -36$

$\implies ab = \frac{-36}{-2}=18\:--(3)$

$Now,\\Value\:of \: a^{3}-b^{3}\\=(a-b)(a^{2}+ab+b^{2})\\=7(85+18)\\=7 \times 103\\=721$

Therefore.,

$Value\:of \: a^{3}-b^{3} = 721$

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