Question

without finding the cubes,factorise and simplify (1/4)^3+(1/3)^3-(7/12)^3

Answer 1

We know that,
(x+y+z)(x²+y²+z²-xy-yz-zx)=x³+y³+z³-3xyz

(x+y+z)(x²+y²+z²-xy-yz-zx)+3xyz=x³+y³+z³
Here x³=(1/4)³ , y³=(1/3)³ , z³=-(7/12)³
Now we have,
(1/4)³+(1/3)³-(7/12)³=(1/4+1/3-7/12)(x²+y²+z²-xy-yz-zx)+3(1/4)(1/3)(-7/12)

(1/4)³+(1/3)³-(7/12)³={(3+4-7)/12}(x²+
y²+z²-xy-yz-zx)-21/144

(1/4)³+(1/3)³-(7/12)³={0/12}³(x²+y²+z²-xy-yz-xz)-21/144

(1/4)³+(1/3)³-(7/12)³=-21/144=-7/48

Hence value of,
(1/4)³+(1/3)³-(7/12)³=-7/48

Answer 2

Answer:

$\frac{-7}{48}$

Step-by-step explanation:

$(1/4)^3+(1/3)^3-(7/12)^3$

To find :

$(1/4)^3+(1/3)^3-(7/12)^3$

Solution:

The left side of the equation must be shown to be equal to the right side in order to establish this relationship. The equation on the left side of the equal to sign in mathematics is referred to as the Left Hand Side, and the equation on the right side is referred to as the Right Hand Side. By extending the right hand side of the provided issue, we may demonstrate that the left and right hands are equal.

The terms in one set of brackets are monomial, whereas they are binomial in the other set. We must multiply each phrase in one bracket by each term in the other bracket in order to solve the right-hand side of the given identity. We also need to multiply the signs by the magnitudes. Positive multiplied with positive yields a positive outcome, while negative multiplied with negative yields a negative result and vice versa.

$(x+y+x)(x^{2} +y^{2} +z^{2} +-xy-yz-zx)=x^{3} +y^{3} +z^{3} -3xy\\x^{3}=(\frac{1}{4} )^{3}\\ y^{3} =(\frac{1}{3} )^{3} \\z^{3} =(\frac{7}{12} )^{3}\\=(\frac{1}{4} )^{3}+(\frac{1}{3} )^{3}-(\frac{7}{12} )^{3}=(\frac{1}{4}+\frac{1}{3}-\frac{7}{12} )(x^{2} +y^{2} +z^{2} +-xy-yz-zx)+3(\frac{1}{4}) (\frac{1}{3})( \frac{-7}{12})\\ (\frac{1}{4} )^{3}+(\frac{1}{3} )^{3}-(\frac{7}{12} )^{3}=({\frac{3+4-7}{12} })^{3}(x^{2} +y^{2} +z^{2}++-xy-yz-zx)-\frac{21}{144}$

$(\frac{1}{4} )^{3}+(\frac{1}{3} )^{3}-(\frac{7}{12} )^{3}=0-\frac{21}{144}\\ (\frac{1}{4} )^{3}+(\frac{1}{3} )^{3}-(\frac{7}{12} )^{3}=\frac{-7}{48}$

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